Bài tập cơ bản bài Quan hệ giữa ba cạnh của một tam giác. Bất đẳng thức tam giác

Home Lớp 7 Toán lớp 7 - Sách cũ Quan hệ giữa ba cạnh của một tam giác. Bất đẳng thức tam giác Bài tập cơ bản

{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"V\u1edbi b\u1ed9 ba \u0111o\u1ea1n th\u1eb3ng c\u00f3 s\u1ed1 \u0111o sau \u0111\u00e2y, b\u1ed9 ba n\u00e0o kh\u00f4ng th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c?","select":[" A. $3cm; 4cm; 5cm$ "," B. $2cm; 4cm; 6cm$ ","C. $6cm; 9cm; 12cm$ ","D. $5cm; 8cm; 10cm$ "],"hint":"S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c \u0111\u1ec3 ki\u1ec3m tra","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Trong tr\u01b0\u1eddng h\u1ee3p x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c $a$ l\u00e0 s\u1ed1 l\u1edbn nh\u1ea5t trong ba s\u1ed1 $a, b, c$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 t\u1ed3n t\u1ea1i tam gi\u00e1c ch\u1ec9 c\u1ea7n: $a <br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> +) Ta c\u00f3: <br\/> $5 < 3 + 4$ <br\/> N\u00ean b\u1ed9 ba s\u1ed1 $3cm; 4cm; 5cm$ c\u00f3 th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <br\/> +) $6 = 2 + 4$ <br\/> N\u00ean b\u1ed9 ba s\u1ed1 $2cm; 4cm; 6cm$ kh\u00f4ng th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <br\/> +) $12 < 6 + 9$ <br\/> N\u00ean b\u1ed9 ba s\u1ed1 $6cm; 9cm; 12cm$ c\u00f3 th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <br\/> +) $10 < 5 + 8$ <br\/> N\u00ean b\u1ed9 ba s\u1ed1 $5cm; 8cm; 10cm$ c\u00f3 th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> <br\/> Nh\u1eadn x\u00e9t: <br\/> +) Khi x\u00e9t \u0111\u1ed9 d\u00e0i ba \u0111o\u1ea1n th\u1eb3ng c\u00f3 th\u1ecfa m\u00e3n b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c kh\u00f4ng ta, ch\u1ec9 c\u1ea7n so s\u00e1nh \u0111\u1ed9 d\u00e0i l\u1edbn nh\u1ea5t v\u1edbi t\u1ed5ng hai \u0111\u1ed9 d\u00e0i c\u00f2n l\u1ea1i <br\/> +) K\u1ebft qu\u1ea3 tr\u00ean c\u00f3 \u0111\u01b0\u1ee3c nh\u1edd v\u00e0o m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p kh\u00e1c \u0111\u1ec3 ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c, c\u1ee5 th\u1ec3: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D07.png' \/><\/center> <br\/> Trong $\\triangle{ABC}$ gi\u1ea3 s\u1eed $BC$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t suy ra $\\widehat{A}$ l\u1edbn nh\u1ea5t <br\/> K\u1ebb $AH \\perp BC$ ($H \\in BC$), ta c\u00f3: <br\/> - Trong $\\triangle{HAB}$ vu\u00f4ng t\u1ea1i $H$, ta c\u00f3: $BH < AB$ (1) <br\/> - Trong $\\triangle{HAC}$ vu\u00f4ng t\u1ea1i $H$, ta c\u00f3: $CH < AC$ (2) <br\/> C\u1ed9ng theo v\u1ebf c\u1ee7a (1) v\u00e0 (2), ta \u0111\u01b0\u1ee3c: <br\/> $BH + CH < AB + AC$ hay $BC < AB + AC$ <br\/> V\u1edbi gi\u1ea3 s\u1eed $BC$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t, n\u00ean c\u00e1c b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f2n l\u1ea1i l\u00e0 hi\u1ec3n nhi\u00ean <\/i> <\/span> ","column":2}]}],"id_ques":1861},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"T\u1ed3n t\u1ea1i m\u1ed9t tam gi\u00e1c c\u00f3 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh l\u00e0 $8m, 12m, 7m$. \u0110\u00fang <\/b> hay sai <\/b>?","select":[" A. \u0110\u00daNG "," B. SAI "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Trong tr\u01b0\u1eddng h\u1ee3p x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c $a$ l\u00e0 s\u1ed1 l\u1edbn nh\u1ea5t trong ba s\u1ed1 $a, b, c$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 t\u1ed3n t\u1ea1i tam gi\u00e1c ch\u1ec9 c\u1ea7n: $a <br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> +) Ta c\u00f3: <br\/> $12 < 7 + 8 $ <br\/> N\u00ean b\u1ed9 ba s\u1ed1 $8m, 12m, 7m$ c\u00f3 th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00daNG <\/span> <br\/> Nh\u1eadn x\u00e9t: Ngo\u00e0i c\u00e1ch tr\u00ean \u0111\u1ec3 x\u00e9t \u0111\u1ed9 d\u00e0i ba \u0111o\u1ea1n th\u1eb3ng c\u00f3 th\u1ecfa m\u00e3n b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c hay kh\u00f4ng ta c\u00f3 th\u1ec3 so s\u00e1nh \u0111\u1ed9 d\u00e0i nh\u1ecf nh\u1ea5t v\u1edbi hi\u1ec7u hai \u0111\u1ed9 d\u00e0i c\u00f2n l\u1ea1i <br\/> Trong tr\u01b0\u1eddng h\u1ee3p x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c $a$ l\u00e0 s\u1ed1 nh\u1ecf nh\u1ea5t trong $3$ s\u1ed1 th\u00ec \u0111i\u1ec1u ki\u1ec7n t\u1ed3n t\u1ea1i tam gi\u00e1c ch\u1ec9 c\u1ea7n $a > |b - c|$ <\/i> <\/span> ","column":2}]}],"id_ques":1862},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110\u1ed9 d\u00e0i hai c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c l\u00e0 $4cm$ v\u00e0 $12cm$. Trong c\u00e1c s\u1ed1 \u0111o sau \u0111\u00e2y, s\u1ed1 \u0111o n\u00e0o l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba c\u1ee7a tam gi\u00e1c \u0111\u00f3? ","select":[" A. $6cm$ "," B. $7cm$ ","C. $8cm$","D. $9cm$"],"hint":"\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho c\u00e1c b\u1ed9 ba s\u1ed1 \u0111o \u0111\u1ed9 d\u00e0i","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Trong tr\u01b0\u1eddng h\u1ee3p x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c $a$ l\u00e0 s\u1ed1 l\u1edbn nh\u1ea5t trong ba s\u1ed1 $a, b, c$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 t\u1ed3n t\u1ea1i tam gi\u00e1c ch\u1ec9 c\u1ea7n: $a <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c, ta c\u00f3: <br\/> +) $12 > 6 + 4 $ <br\/> N\u00ean $6cm$ kh\u00f4ng th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba c\u1ee7a tam gi\u00e1c <br\/> +) $12 > 4 + 7$ <br\/> N\u00ean $7cm$ kh\u00f4ng th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba c\u1ee7a tam gi\u00e1c <br\/> +) $12 = 8 + 4$ <br\/> N\u00ean $8cm$ kh\u00f4ng th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba c\u1ee7a tam gi\u00e1c <br\/> +) $12 < 9 + 4$ <br\/> N\u00ean $9cm$ c\u00f3 th\u1ec3 l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba c\u1ee7a tam gi\u00e1c <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1863},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["37"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" Bi\u1ebft hai c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c c\u00e2n l\u00e0 $7cm$ v\u00e0 $15cm$. T\u00ednh chu vi c\u1ee7a tam gi\u00e1c. <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_ $cm$ ","hint":"D\u1ef1a v\u00e0o b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c \u0111\u1ec3 t\u00ecm c\u1ea1nh th\u1ee9 ba c\u1ee7a tam gi\u00e1c c\u00e2n","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1:<\/b> T\u00ecm c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a tam gi\u00e1c c\u00e2n (b\u1eb1ng m\u1ed9t trong hai c\u1ea1nh kia) <br\/>B\u01b0\u1edbc 2:<\/b> T\u00ednh chu vi c\u1ee7a tam gi\u00e1c \u0111\u00f3 <br\/> <br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> V\u00ec tam gi\u00e1c \u0111\u00e3 cho l\u00e0 tam gi\u00e1c c\u00e2n n\u00ean \u0111\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba l\u00e0 $7cm$ ho\u1eb7c $15cm$ <br\/> \u0110\u1ed9 d\u00e0i c\u1ea1nh th\u1ee9 ba kh\u00f4ng th\u1ec3 l\u00e0 7cm v\u00ec $7 + 7 < 15$ <br\/> Tr\u01b0\u1eddng h\u1ee3p c\u1ea1nh th\u1ee9 ba l\u00e0 $15cm$ th\u1ecfa m\u00e3n b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00ec $15 < 15 + 7$ <br\/> Chu vi c\u1ee7a tam gi\u00e1c l\u00e0: <br\/> $15 + 15 + 7 = 37 (cm)$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $37$ <\/span> "}]}],"id_ques":1864},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"T\u1ed3n t\u1ea1i m\u1ed9t tam gi\u00e1c c\u00f3 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh l\u00e0 $a, b, c$ sao cho $a = \\dfrac{3}{2}b, b = \\dfrac{3}{2}c$. \u0110\u00fang <\/b> hay sai <\/b>?","select":[" A. \u0110\u00daNG "," B. SAI "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ednh $c$ theo $b$ r\u1ed3i \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho 3 s\u1ed1 xem c\u00f3 th\u1ecfa m\u00e3n kh\u00f4ng <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $b = \\dfrac{3}{2}c \\Rightarrow c = b : \\dfrac{3}{2} = \\dfrac{2}{3}b$ <br\/> X\u00e9t b\u1ed9 ba s\u1ed1 $\\dfrac{3}{2}b; b; \\dfrac{2}{3}b$ <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c, ta c\u00f3: <br\/> $\\dfrac{3}{2}b N\u00ean b\u1ed9 ba s\u1ed1 $a = \\dfrac{3}{2}b; b = \\dfrac{3}{2}c; c$ c\u00f3 th\u1ec3 l\u00e0 ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1865},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" Tam gi\u00e1c $ABC$ c\u00f3 $AB = 1dm$, $AC = 5dm$, \u0111\u1ed9 d\u00e0i $BC$ (t\u00ednh b\u1eb1ng \u0111\u1ec1-xi-m\u00e9t) l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean. T\u00ednh \u0111\u1ed9 d\u00e0i $BC$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $BC$ = _input_ $dm$ ","hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c so s\u00e1nh hi\u1ec7u \u0111\u1ed9 d\u00e0i hai c\u1ea1nh v\u1edbi t\u1ed5ng \u0111\u1ed9 d\u00e0i hai c\u1ea1nh \u0111\u1ec3 t\u00ecm \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u00f2n l\u1ea1i <br\/> <br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho tam gi\u00e1c $ABC$, ta c\u00f3: <br\/> $AC - AB < BC < AC + AB$ <br\/> $\\Leftrightarrow$ $5 - 1 < BC < 5 + 1$ <br\/> $\\Leftrightarrow 4 < BC < 6$ <br\/> V\u00ec \u0111\u1ed9 d\u00e0i $BC$ l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean (\u0111\u1ec1-xi-m\u00e9t) <br\/> N\u00ean $BC = 5dm$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $5$ <\/span> "}]}],"id_ques":1866},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["17"],["19"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" Tam gi\u00e1c $MNP$ c\u00f3 $MN = 3cm$, $NP = 19cm$, \u0111\u1ed9 d\u00e0i $MP$ (t\u00ednh b\u1eb1ng x\u0103ng-ti-m\u00e9t) l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean t\u1ed1. T\u00ednh \u0111\u1ed9 d\u00e0i $MP$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $MP$ = _input_ $cm$ ho\u1eb7c $MP$ = _input_ $cm$ ","hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c so s\u00e1nh hi\u1ec7u \u0111\u1ed9 d\u00e0i hai c\u1ea1nh v\u1edbi t\u1ed5ng \u0111\u1ed9 d\u00e0i hai c\u1ea1nh \u0111\u1ec3 t\u00ecm \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u00f2n l\u1ea1i <br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho tam gi\u00e1c $MNP$, ta c\u00f3: <br\/> $NP - MN < MP < NP + MN$ <br\/> $\\Leftrightarrow$ $19 - 3 < MP < 19 + 3$ <br\/> $\\Leftrightarrow 16 < MP < 22$ <br\/> V\u00ec \u0111\u1ed9 d\u00e0i $MP$ l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean t\u1ed1 (x\u0103ng-xi-m\u00e9t) <br\/> N\u00ean $MP = 17cm$ ho\u1eb7c $MP = 19cm$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $17; 19$ <\/span> "}]}],"id_ques":1867},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho $\\triangle{ABC}$ c\u00f3 $\\widehat{A} = \\widehat{B} = 40^{o}$. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? ","select":[" A. $AB = AC > BC$ "," B. $AB > AC = BC$ ","C. $CA = CB > AB$ ","D. $AB = AC < BC$ "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> S\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00fd t\u1ed5ng ba g\u00f3c \u0111\u1ec3 t\u00ednh s\u1ed1 \u0111o g\u00f3c $C$ <br\/> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh g\u00f3c $C$ v\u1edbi g\u00f3c $A$ v\u00e0 g\u00f3c $B$ <br\/> B\u01b0\u1edbc 3: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n \u0111\u1ec3 so s\u00e1nh \u0111\u1ed9 d\u00e0i ba c\u1ea1nh <\/span> <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> $\\triangle{ABC}$ c\u00f3: <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{C} = 180^{o} - 40^{o} - 40^{o} = 100^{o}$ <br\/> V\u00ec $100^{o} > 40^{o} = 40^{o}$ n\u00ean $\\widehat{C} > \\widehat{A} = \\widehat{B}$ <br\/> $\\Rightarrow$ $AB > BC = AC$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{ABC}$) <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}]}],"id_ques":1868},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho $\\triangle{ABC}$ c\u00f3 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean: $AB = 5cm; BC = 4cm$. Chu vi tam gi\u00e1c $ABC$ kh\u00f4ng th\u1ec3 c\u00f3 s\u1ed1 \u0111o n\u00e0o sau \u0111\u00e2y? ","select":["A. $18cm$ ","B. $15cm$ ","C. $12cm$ ","D. $17cm$ "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c $AB - BC < AC < AB + BC$ \u0111\u1ec3 t\u00ecm \u0111\u1ed9 d\u00e0i c\u1ea1nh $AC$ <br\/> B\u01b0\u1edbc 2: <\/b> T\u00ednh chu vi tam gi\u00e1c $ABC$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho tam gi\u00e1c $ABC$, ta c\u00f3: <br\/> $AB - BC < AC < AB + BC$ <br\/> $\\Leftrightarrow 5 - 4 < AC < 5 + 4$ <br\/> $\\Leftrightarrow 1 < AC < 9$ (1) <br\/> Chu vi tam gi\u00e1c $ABC$ l\u00e0: <br\/> $AC + AB + BC = AC + 5 + 4 = AC + 9$ (cm) <br\/> K\u1ebft h\u1ee3p v\u1edbi (1) ta c\u00f3: <br\/> $10 < AC + 9 < 18$ m\u00e0 $AC < 9$ <br\/> N\u00ean chu vi tam gi\u00e1c $ABC$ kh\u00f4ng th\u1ec3 l\u00e0 $18cm$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":2}]}],"id_ques":1869},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho $\\triangle{ABC}$. G\u1ecdi $M$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec n\u1eb1m trong tam gi\u00e1c \u0111\u00f3. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $MA + MB + MC > \\dfrac{AB + BC + CA}{2} $ ","B. $MA + MB + MC = \\dfrac{AB + BC + CA}{2}$","C. $MA + MB + MC > AB + BC + CA$","D. $MA + MB + MC = AB + BC + CA$ "],"hint":"S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam v\u1edbi c\u00e1c tam gi\u00e1c $AMB; AMC; BMC$ <br\/> B\u01b0\u1edbc 2: <\/b> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a c\u00e1c b\u1ea5t \u0111\u1eb3ng th\u1ee9c l\u1ea1i \u0111\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec1u c\u1ea7n t\u00ecm. <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c: <br\/> - V\u1edbi $\\triangle{AMB}$ ta c\u00f3: $MA + MB > AB$ (1) <br\/> - V\u1edbi $\\triangle{AMC}$ ta c\u00f3: $MA + MC > AC$ (2) <br\/> - V\u1edbi $\\triangle{BMC}$ ta c\u00f3: $MB + MC > BC$ (3) <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c (1), (2), (3), ta \u0111\u01b0\u1ee3c: <br\/> $2(MA + MB + MC) > AB + AC + BC$ <br\/> Hay $MA + MB + MC > \\dfrac{AB + BC + AC}{2}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":1}]}],"id_ques":1870},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho $\\triangle{MNP}$, \u0111i\u1ec3m $I$ n\u1eb1m trong tam gi\u00e1c. <br\/> Khi \u0111\u00f3 $IM + IN < MP + NP$. \u0110\u00fang <\/b> hay sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MI$ v\u1edbi $NP$. \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho c\u00e1c tam gi\u00e1c $MHP$ v\u00e0 $INH$","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $MI$ v\u1edbi $NP$. \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho c\u00e1c tam gi\u00e1c $MHP$ v\u00e0 $INH$ <br\/> B\u01b0\u1edbc 2: <\/b> C\u1ed9ng c\u00e1c v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u1edf b\u01b0\u1edbc 1 l\u1ea1i \u0111\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec1u c\u1ea7n t\u00ecm. <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D01.png' \/><\/center> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c: <br\/> - V\u1edbi $\\triangle{MHP}$ ta c\u00f3: $MH < HP + MP$ (1) <br\/> - V\u1edbi $\\triangle{INH}$ ta c\u00f3: $IN < IH + NH$ (2) <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a \u0111\u1eb3ng th\u1ee9c (1), (2) ta \u0111\u01b0\u1ee3c: <br\/> $MH + IN < HP + MP + IH + NH$ <br\/> Hay $MI + IH + IN < MP + (HP + NH) + IH$ <br\/> $ \\Leftrightarrow MI + IH + IN < MP + NP + IH$ <br\/> $\\Leftrightarrow MI + IN < MP + NP$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $MI + IN < MP + NP$ l\u00e0 \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1871},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["16"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" \u0110\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c t\u1ec9 l\u1ec7 v\u1edbi c\u00e1c s\u1ed1 $2; 3; 4$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh l\u1edbn nh\u1ea5t bi\u1ebft t\u1ed5ng \u0111\u1ed9 d\u00e0i hai c\u1ea1nh kia l\u00e0 $20$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_ ","hint":"S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ee7a d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau","explain":" G\u1ecdi $3$ c\u1ea1nh c\u1ee7a tam gi\u00e1c l\u1ea7n l\u01b0\u1ee3t l\u00e0 $a; b; c$ <br\/> Kh\u00f4ng m\u1ea5t t\u00ednh t\u1ed5ng qu\u00e1t, gi\u1ea3 s\u1eed $a \\leq b \\leq c$ <br\/> T\u1eeb \u0111\u1ec1 b\u00e0i ta c\u00f3: $\\dfrac{a}{2} = \\dfrac{b}{3} = \\dfrac{c}{4}$ v\u00e0 $a + b = 20$ <br\/> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: <br\/> $\\dfrac{a}{2} = \\dfrac{b}{3} = \\dfrac{c}{4} = \\dfrac{a + b}{2 + 3} = \\dfrac{20}{2 + 3} = 4$ <br\/> $\\Rightarrow$ $c = 4 . 4 = 16$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $16$ <\/span> "}]}],"id_ques":1872},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["42"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" \u0110\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c t\u1ec9 l\u1ec7 v\u1edbi c\u00e1c s\u1ed1 $2; 3; 4$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh nh\u1ecf nh\u1ea5t bi\u1ebft hi\u1ec7u \u0111\u1ed9 d\u00e0i hai c\u1ea1nh kia l\u00e0 $21cm$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_$cm$ ","hint":"S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ee7a d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau","explain":" G\u1ecdi $3$ c\u1ea1nh c\u1ee7a tam gi\u00e1c l\u1ea7n l\u01b0\u1ee3t l\u00e0 $a; b; c$ <br\/> Kh\u00f4ng m\u1ea5t t\u00ednh t\u1ed5ng qu\u00e1t, gi\u1ea3 s\u1eed $a \\leq b \\leq c$ <br\/> T\u1eeb \u0111\u1ec1 b\u00e0i ta c\u00f3: $\\dfrac{a}{2} = \\dfrac{b}{3} = \\dfrac{c}{4}$ v\u00e0 $c - b = 21$ <br\/> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: <br\/> $\\dfrac{a}{2} = \\dfrac{b}{3} = \\dfrac{c}{4} = \\dfrac{c - b}{4 - 3} = \\dfrac{21}{1} = 21$ <br\/> $\\Rightarrow$ $a = 21 . 2 = 42(cm)$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $42$ <\/span> "}]}],"id_ques":1873},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["14"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" Cho tam gi\u00e1c c\u00e2n c\u00f3 chu vi b\u1eb1ng $34cm$ v\u00e0 m\u1ed9t c\u1ea1nh c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng $6cm$. T\u00ednh \u0111\u1ed9 d\u00e0i hai c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a tam gi\u00e1c \u0111\u00f3. <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_ $cm$ ","hint":"X\u00e9t hai kh\u1ea3 n\u0103ng: C\u1ea1nh \u0111\u00e3 cho l\u00e0 c\u1ea1nh \u0111\u00e1y ho\u1eb7c c\u1ea1nh b\u00ean","explain":"V\u00ec tam gi\u00e1c \u0111\u00e3 cho l\u00e0 tam gi\u00e1c c\u00e2n n\u00ean hai c\u1ea1nh b\u00ean b\u1eb1ng nhau <br\/> N\u1ebfu c\u1ea1nh b\u00ean d\u00e0i $6cm$, th\u00ec do chu vi b\u1eb1ng $34cm$, n\u00ean c\u1ea1nh \u0111\u00e1y d\u00e0i l\u00e0: <br\/> $34 - 6 . 2 = 22(cm)$ <br\/> V\u00ec $22 > 6 + 6$ (kh\u00f4ng th\u1ecfa m\u00e3n b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> N\u1ebfu c\u1ea1nh \u0111\u00e1y b\u1eb1ng $6cm$, do chu vi b\u1eb1ng $34cm$, n\u00ean c\u1ea1nh b\u00ean d\u00e0i l\u00e0: <br\/> $(34 - 6) : 2 = 14(cm)$ <br\/> V\u00ec $14 < 14 + 6$ (th\u1ecfa m\u00e3n b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> N\u00ean \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u1ea7n t\u00ecm l\u00e0 $14cm$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $14$ <\/span> "}]}],"id_ques":1874},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho g\u00f3c nh\u1ecdn $xOy$. Tr\u00ean tia $Ox$ l\u1ea5y hai \u0111i\u1ec3m $A$, $B$ sao cho $A$ n\u1eb1m gi\u1eefa $O$ v\u00e0 $B$. Tr\u00ean c\u1ea1nh $Oy$ l\u1ea5y hai \u0111i\u1ec3m $C$, $D$ sao cho $C$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $O$ v\u00e0 $D$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AB + CD < AD + BC$ ","B. $AB + CD = AD + BC$","C. $AB + CD > AD + BC$"],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BD$ <br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi c\u00e1c tam gi\u00e1c $AIB; CID$ <br\/> B\u01b0\u1edbc 2: <\/b> C\u1ed9ng c\u00e1c v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c l\u1ea1i \u0111\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec1u c\u1ea7n t\u00ecm. <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D02.png' \/><\/center> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BD$ <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c: <br\/> - V\u1edbi $\\triangle{AIB}$ ta c\u00f3: $AB < IA + IB$ (1) <br\/> - V\u1edbi $\\triangle{CID}$ ta c\u00f3: $CD < IC + ID$ (2) <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c (1), (2) ta \u0111\u01b0\u1ee3c: <br\/> $AB + CD < IA + IB + IC + ID$ <br\/> Hay $AB + CD < AD + BC$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":1}]}],"id_ques":1875},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} > 90^{o}$, $AB = \\dfrac{1}{2}AC$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $BC > AB$ ","B. $BC = AB$","C. $BC < AB$"],"hint":"","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D03.png' \/><\/center> Tam gi\u00e1c $ABC$ c\u00f3: $AB = \\dfrac{1}{2}AC$ (gt) <br\/> $\\widehat{B} > 90^{o}$ $\\Rightarrow$ $AC$ l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho tam gi\u00e1c $ABC$, ta c\u00f3: <br\/> $BC > AC - AB$ <br\/> Hay $BC > 2AB - AB = AB$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":3}]}],"id_ques":1876},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","f","t"]],"list":[{"point":5,"image":"","col_name":["Cho $\\triangle{ABC}$ v\u00e0 $M$ l\u00e0 m\u1ed9t \u0111i\u1ec3m n\u1eb1m trong tam gi\u00e1c. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $BM$ v\u00e0 c\u1ea1nh $AC$ ","\u0110\u00fang","Sai"],"arr_ques":["$MA < MI + AI$ <br\/> $MA + MB < IA + IB$ ","$IB < IC + CB$ "," $IB + IA > CA + CB$ "," $MA + MB < CA + CB$ "],"hint":"\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho c\u00e1c tam gi\u00e1c $AMI; BIC$ ","explain":[" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D04.png' \/><\/center> <br\/>\u0110\u00daNG v\u00ec: <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{AMI}$, ta c\u00f3: <br\/> $MA < MI + IA$ <br\/> $\\Rightarrow$ $MA + MB < MI + IA + MB$ <br\/> $\\Rightarrow$ $MA + MB < IA + IB$ "," <br\/> \u0110\u00daNG v\u00ec: <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{BIC}$ ta c\u00f3: <br\/> $IB < IC + CB$ ","<br\/> SAI v\u00ec: <br\/> Ta c\u00f3: $IB < IC + CB$ (ch\u1ee9ng minh tr\u00ean) <br\/> N\u00ean: $IB + IA < IC + CB + IA$ <br\/> $\\Rightarrow$ $IB + IA < CA + CB$ ","<br\/> \u0110\u00daNG v\u00ec: <br\/> Ta c\u00f3: <br\/> $MA + MB < IA + IB$ (cmt) <br\/> $IB + IA < CA + CB$ (cmt) <br\/> $\\Rightarrow$ $MA + MB < CA + CB$ "]}]}],"id_ques":1877},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"\u0110\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a m\u1ed9t tam gi\u00e1c nh\u1ecf h\u01a1n n\u1eeda chu vi c\u1ee7a tam gi\u00e1c \u1ea5y. \u0110\u00fang <\/b> hay sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI"],"hint":"","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> V\u1ebd \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$ c\u1ee7a $\\triangle{ABC}$ <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{AMB}$ v\u00e0 $\\triangle{AMC}$ <br\/> B\u01b0\u1edbc 2: <\/b> C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u1ee7a hai b\u1ea5t \u0111\u1eb3ng th\u1ee9c tr\u00ean l\u00e0m xu\u1ea5t hi\u1ec7n chu vi tam gi\u00e1c <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D05.png' \/><\/center> <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{AMB}$, ta c\u00f3: <br\/> $AM < AB + BM$ (1) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{AMC}$, ta c\u00f3: <br\/> $AM < AC + MC$ (2) <br\/> C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u1ee7a (1) v\u00e0 (2), ta \u0111\u01b0\u1ee3c: <br\/> $2AM < AB + BM + AC + CM$ <br\/> $\\Rightarrow$ $AM < \\dfrac{1}{2}(AB + AC + BC)$ <br\/> Hay \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung tuy\u1ebfn nh\u1ecf h\u01a1n n\u1eeda chu vi c\u1ee7a tam gi\u00e1c <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1878},{"time":24,"part":[{"title":"Ch\u1ecdn nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3"]],"list":[{"point":10,"img":"","ques":"Cho tam gi\u00e1c v\u1edbi \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh l\u1ea7n l\u01b0\u1ee3t l\u00e0 $a, b, c$; bi\u1ebft $a \\geq b \\geq c$. <br\/> Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang?","hint":"\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c","column":2,"number_true":2,"select":["A. $a < \\dfrac{a + b + c}{2}$ ","B. $a > \\dfrac{a + b + c}{2}$ ","C. $a \\geq \\dfrac{a + b + c}{3}$ ","D. $a < \\dfrac{a + b + c}{3}$ "],"explain":"+) Theo \u0111\u1ec1 b\u00e0i: $a \\geq b \\geq c$<br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c ta c\u00f3: <br\/> $a C\u1ed9ng $a$ v\u00e0 hai v\u1ebf ta \u0111\u01b0\u1ee3c: $a + a < a + b + c$ <br\/> $\\Rightarrow$ $a < \\dfrac{a + b + c}{2}$ <br\/> +) V\u00ec $a \\geq b \\geq c$ n\u00ean $2a \\geq b + c$ <br\/> C\u1ed9ng $a$ v\u00e0o hai v\u1ebf ta \u0111\u01b0\u1ee3c: $3a \\geq a + b + c$ <br\/> $\\Rightarrow$ $a \\geq \\dfrac{a + b + c}{3}$ <br\/> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A, C "}]}],"id_ques":1879},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho tam gi\u00e1c $MNP$. G\u1ecdi $I$ l\u00e0 m\u1ed9t \u0111i\u1ec3m thu\u1ed9c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $M$ ($I$ kh\u00f4ng tr\u00f9ng v\u1edbi $M$). Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $MN + MP > IP + IN$ ","B. $MN + MP < IP + IN$","C. $MN + MP = IP + IN$","D. M\u1ed9t k\u1ebft qu\u1ea3 kh\u00e1c"],"hint":"\u0110\u1ec3 xu\u1ea5t hi\u1ec7n t\u1ed5ng $MN + MP$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MN$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MP$, khi \u0111\u00f3 $MN + MP = ND$","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MN$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MP$ <br\/> Khi \u0111\u00f3 $MN + MP = ND$ <br\/> Ch\u1ee9ng minh $ID = IP$ <br\/> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{NID}$ <br\/> Sau \u0111\u00f3 bi\u1ebfn \u0111\u1ed5i l\u00e0m xu\u1ea5t hi\u1ec7n $IP + IN$ v\u00e0 $MN + MP$ \u0111\u1ec3 so s\u00e1nh <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv1/img\/H7C3B19_D06.png' \/><\/center> <br\/> \u0110\u1ec3 xu\u1ea5t hi\u1ec7n t\u1ed5ng $MN + MP$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MN$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MP$ <br\/> Ta c\u00f3: $MN + MD = ND$ hay $MN + MP = ND$ (1) <br\/> X\u00e9t $\\triangle{MID}$ v\u00e0 $\\triangle{MIP}$ c\u00f3: <br\/> $\\begin{cases} MI \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{DMI} = \\widehat{PMI} (gt) \\\\ MP = MD (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} D) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{MIP} = \\triangle{MID}$ (c.g.c) <br\/> $\\Rightarrow$ $IP = ID$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{NID}$, ta c\u00f3: <br\/> $ND < ID + IN = IP + IN$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra: $MN + MP < IP + IN$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}]}],"id_ques":1880}],"lesson":{"save":0,"level":1}}

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Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

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