{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 hai \u0111i\u1ec3m $A$, $B$ n\u1eb1m v\u1ec1 hai ph\u00eda c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $d$. V\u1ecb tr\u00ed \u0111i\u1ec3m $C$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $d$ sao cho $CA + CB$ nh\u1ecf nh\u1ea5t l\u00e0: ","select":["A. $AC \\perp d$ ","B. $BC \\perp d$ ","C. $AB \\cap d = \\left\\{ C \\right\\}$ ","D. C\u1ea3 $3$ \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u sai"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K06.png' \/><\/center> <br\/> G\u1ecdi $C_{0}$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $d$<br\/> $C$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $d$ <br\/> Trong $\\triangle{ABC}$, ta c\u00f3: <br\/> $AB < AC + BC$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $AC_{0} + C_{0}B = BC$ <br\/> $\\Leftrightarrow$ $AC_{0} + C_{0}B < AC + BC$ <br\/> V\u1eady \u0111\u1ec3 $CA + CB$ nh\u1ecf nh\u1ea5t th\u00ec $C \\equiv C_{0}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: B\u00e0i to\u00e1n tr\u00ean minh h\u1ecda ph\u01b0\u01a1ng ph\u00e1p t\u00ecm \u0111i\u1ec3m $C$ \u0111\u1ec3 $CA + CB$ nh\u1ecf nh\u1ea5t, trong tr\u01b0\u1eddng h\u1ee3p $A$, $B$ n\u1eb1m kh\u00e1c ph\u00eda \u0111\u1ed1i v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $d$ <\/span> ","column":2}]}],"id_ques":1881},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 hai \u0111i\u1ec3m $A$, $B$ n\u1eb1m v\u1ec1 m\u1ed9t ph\u00eda c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 $AB$ kh\u00f4ng song song v\u1edbi $d$. M\u1ed9t \u0111i\u1ec3m $M$ di \u0111\u1ed9ng tr\u00ean $d$. V\u1ecb tr\u00ed \u0111i\u1ec3m $M$ \u0111\u1ec3 $|MA - MB|$ l\u1edbn nh\u1ea5t l\u00e0: ","select":["A. $\\left\\{ M \\right\\} = AB \\cap d $ ","B. $BM \\perp d$ ","C. $AM \\perp d $ ","D. $IM \\perp d; IA = IB, I \\in AB$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K07.png' \/><\/center> <br\/> V\u00ec $AB$ kh\u00f4ng song song v\u1edbi $d$ n\u00ean g\u1ecdi $N$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $d$, $M$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean $d$ <br\/> V\u1edbi \u0111i\u1ec3m $M$ b\u1ea5t k\u00ec thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $d$, $M$ kh\u00f4ng tr\u00f9ng v\u1edbi $N$ <br\/> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{MAB}$, ta c\u00f3: <br\/> $|MA - MB| < AB$ (1) <br\/> Khi $M \\equiv N$ th\u00ec $|MA - MB| = AB$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $|MA - MB|$ l\u1edbn nh\u1ea5t l\u00e0 b\u1eb1ng $AB$ khi $M \\equiv N$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $d$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: B\u00e0i to\u00e1n tr\u00ean minh h\u1ecda ph\u01b0\u01a1ng ph\u00e1p t\u00ecm \u0111i\u1ec3m $M$ \u0111\u1ec3 $MA - MB$ l\u1edbn nh\u1ea5t, trong tr\u01b0\u1eddng h\u1ee3p $A$, $B$ n\u1eb1m c\u00f9ng ph\u00eda \u0111\u1ed1i v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $d$ <\/span> ","column":2}]}],"id_ques":1882},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, tia ph\u00e2n gi\u00e1c g\u00f3c $B$ c\u1eaft $AC$ t\u1ea1i $D$. H\u00e3y so s\u00e1nh $BC - BA$ v\u1edbi $DC - DA$. ","select":["A. $BC - BA = DC - DA$ ","B. $BC - BA > DC - DA$ ","C. $BC - BA < DC - DA$ "],"hint":"\u0110\u1ec3 l\u00e0m xu\u1ea5t hi\u1ec7n hi\u1ec7u $BC - BA$, ta k\u1ebb $DH \\perp BC$ ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> K\u1ebb $DH \\perp BC$ <br\/> Ch\u1ee9ng minh $BA = BH$; $DA = DH$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $DC$ v\u00e0 $HC$ v\u00e0 $DC - DH$ v\u00e0 $HC$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u1eeb b\u01b0\u1edbc $1$ v\u00e0 b\u01b0\u1edbc $2$, bi\u1ebfn \u0111\u1ed5i v\u00e0 so s\u00e1nh $BC - BA$ v\u1edbi $DC - DA$ <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K08.png' \/><\/center> <br\/> $\\blacktriangleright$ \u0110\u1ec3 l\u00e0m xu\u1ea5t hi\u1ec7n hi\u1ec7u $BC - BA$, ta k\u1ebb $DH \\perp BC$ <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $BAD$ v\u00e0 $BHD$ c\u00f3: <br\/> $\\begin{cases} \\widehat{B_{1}} = \\widehat{B_{2}} (gt) \\\\ BD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{BAD} = \\triangle{BHD}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow$ $BA = BH$ v\u00e0 $DA = DH$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\blacktriangleright$ $\\triangle{DHC}$ c\u00f3 $DC > DH$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c) <br\/> $HC > DC - DH$ (h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) (1) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $HC = BC - BH = BC - BA$ (v\u00ec $BA = BH$) (2) <br\/> $DC - DH = DC - DA$ (v\u00ec $DH = DA$ (3) <br\/> $\\blacktriangleright$ T\u1eeb (1), (2), (3) $\\Rightarrow$ $BC - BA > DC - DA$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":1}]}],"id_ques":1883},{"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":[[["8"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":" Tam gi\u00e1c $ABC$ c\u00f3 chu vi $18cm$, $BC > AC > AB$. T\u00ednh \u0111\u1ed9 d\u00e0i $BC$, bi\u1ebft r\u1eb1ng \u0111\u1ed9 d\u00e0i \u0111\u00f3 l\u00e0 m\u1ed9t s\u1ed1 ch\u1eb5n (\u0111\u01a1n v\u1ecb: $cm$) <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $BC$ = _input_ $cm$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh $BC + BC + BC$ v\u1edbi $AB + AC + BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $BC$ v\u1edbi $AB + AC$, bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 so s\u00e1nh $BC + BC$ v\u1edbi $AB + AC + BC$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> K\u1ebft h\u1ee3p \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec1 b\u00e0i \u0111\u1ec3 t\u00ecm $BC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> Ta c\u00f3: $BC > AC$; $BC > AB$ <br\/> N\u00ean: $BC + BC + BC > AB + BC + AC$ <br\/> Hay $3BC > 18$, v\u1eady $BC > 6(cm)$ (1) <br\/> M\u1eb7t kh\u00e1c: $BC < AC + AB$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> N\u00ean $BC + BC < AB + AC + BC$ (c\u1ed9ng th\u00eam $BC$ v\u00e0o hai v\u1ebf) <br\/> Hay $2BC < 18$, n\u00ean $BC < 9(cm)$ (2) <br\/> Do $BC$ l\u00e0 m\u1ed9t s\u1ed1 ch\u1eb5n, n\u00ean t\u1eeb (1) v\u00e0 (2) suy ra $BC = 8cm$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: $8$ <\/span> "}]}],"id_ques":1884},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"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. $\\widehat{A} < 2\\widehat{C} $ ","B. $\\widehat{A} = 2\\widehat{C} $ ","C. $\\widehat{A} > 2\\widehat{C}$ "],"hint":"V\u1ebd \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n $AC$ c\u1eaft $BC$ t\u1ea1i $D$. So s\u00e1nh $\\widehat{A}$ v\u1edbi $2\\widehat{MAD}$","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K01.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $AC$ <br\/> V\u1ebd \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ c\u1eaft c\u1ea1nh $BC$ t\u1ea1i $D$ <br\/> X\u00e9t $\\triangle{ABC}$ c\u00f3: $BC > AC - AB = AB$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> N\u00ean \u0111i\u1ec3m $D$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $B$ v\u00e0 $C$ <br\/> Ta c\u00f3: $MA = MC$ ($M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$) <br\/> $AB = \\dfrac{1}{2}AC$ (gt) <br\/> $\\Rightarrow$ $AM = AB$ <br\/> $\\Rightarrow$ $\\triangle{AMB}$ c\u00e2n t\u1ea1i $A$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $\\widehat{ABM} = \\widehat{AMB}$ <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c, $\\widehat{ABC} > 90^{o}$ (gt) <br\/> $\\widehat{AMD} = 90^{o}$ ($MD$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AC$) <br\/> $\\Rightarrow$ $\\widehat{ABC} > \\widehat{AMD}$ <br\/> $\\Rightarrow$ $\\widehat{ABC} - \\widehat{ABM} > \\widehat{AMD} - \\widehat{AMB}$ (c\u00f9ng tr\u1eeb \u0111i $\\widehat{ABM}$) <br\/> $\\Rightarrow$ $\\widehat{MBC} > \\widehat{BMD}$ hay $\\widehat{MBD} > \\widehat{BMD}$ <br\/> $\\Rightarrow MD > BD$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong $\\triangle{BMD}$) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABD}$ v\u00e0 $\\triangle{AMD}$ c\u00f3: <br\/> $\\begin{cases} AB = AM (cmt) \\\\ BD < MD (cmt) \\\\ AD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{BAD} < \\widehat{MAD}$ (\u0111\u1ecbnh l\u00fd b\u1ed5 sung v\u1ec1 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n) <br\/> $\\Rightarrow$ $\\widehat{BAD} + \\widehat{MAD} < \\widehat{MAD} + \\widehat{MAD}$ <br\/> $\\Leftrightarrow \\widehat{A} < 2.\\widehat{MAD}$ (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{DAM}$ v\u00e0 $\\triangle{DCM}$ c\u00f3: <br\/> $\\begin{cases} AM = MC (cmt) \\\\ \\widehat{MAD} = \\widehat{DMC} = 90^{o} (MD \\perp AC) \\\\ DM \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{DAM} = \\triangle{DCM}$ (c.g.c) <br\/> $\\Rightarrow$ $\\widehat{MAD} = \\widehat{C}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{A} < 2\\widehat{C}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: \u1ede trong b\u00e0i n\u00e0y ta \u0111\u00e3 d\u00f9ng \u0111\u1ecbnh l\u00fd b\u1ed5 sung v\u1ec1 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c \u0111\u1ec3 so s\u00e1nh g\u00f3c. (N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau nh\u01b0ng g\u00f3c xen gi\u1eef kh\u00f4ng b\u1eb1ng nhau. C\u1ee5 th\u1ec3 c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n l\u00e0 c\u1ea1nh l\u1edbn h\u01a1n v\u00e0 ng\u01b0\u1ee3c l\u1ea1i) <\/i> <\/span> ","column":3}]}],"id_ques":1885},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $AB > AC$, $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $BAC$, $M$ l\u00e0 \u0111i\u1ec3m thu\u1ed9c \u0111o\u1ea1n th\u1eb3ng $AD$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang?","select":["A. $MB - MC < AB - AC $ ","B. $MB - MC = AB - AC $ ","C. $MB - MC > AB - AC$ "],"hint":"\u0110\u1ec3 l\u00e0m xu\u1ea5t hi\u1ec7n hi\u1ec7u $AB - AC$, tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $AC = AK$, khi \u0111\u00f3 $BK = AB - AC$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $AC = AK$ <br\/> Ch\u1ee9ng minh $MK = MC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{MKB}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $MB - MC$ v\u1edbi $AB - AC$ <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K02.png' \/><\/center> <br\/> $\\blacktriangleright$ Theo gi\u1ea3 thi\u1ebft $AB > AC$, n\u00ean tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $AK = AC$ <br\/> Ta c\u00f3: $BK = AB - AK = AB - AC$ (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMK}$ v\u00e0 $\\triangle{AMC}$ c\u00f3: <br\/> $\\begin{cases} AM \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{A_{1}} = \\widehat{A_{2}} (gt) \\\\ AK = AC (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} K) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMK} = \\triangle{AMC}$ (c.g.c) <br\/> $\\Rightarrow$ $MK = MC$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{MKB}$, ta c\u00f3: <br\/> $MB - MK < BK$ (3) <br\/> T\u1eeb (1), (2), (3), ta c\u00f3: $MB - MC < AB - AC$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: \u0110\u1ec3 l\u00e0m xu\u1ea5t hi\u1ec7n hi\u1ec7u \u0111\u1ed9 d\u00e0i hai c\u1ea1nh ta \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c <\/i> <\/span> ","column":1}]}],"id_ques":1886},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$, $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $BC$. <br\/> Khi \u0111\u00f3: $\\dfrac{AB + AC - BC}{2} < AM < \\dfrac{AB + AC}{2}$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MA$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{MAB}$ v\u00e0 $\\triangle{MAC}$ bi\u1ebfn \u0111\u1ed5i l\u00e0m xu\u1ea5t hi\u1ec7n $AB + AC - BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MA$ <br\/> Ch\u1ee9ng minh $AB = CD$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{ACD}$ sau \u0111\u00f3 bi\u1ebfn \u0111\u1ed5i l\u00e0m xu\u1ea5t hi\u1ec7n t\u1ed5ng $AB + AC$ <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K03.png' \/><\/center> <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c: <br\/> - V\u1edbi $\\triangle{MAB}$, ta c\u00f3: $AM > AB - BM$ (1) <br\/> - V\u1edbi $\\triangle{MAC}$, ta c\u00f3: $AM > AC - MC$ (2) <br\/> C\u1ed9ng t\u1eebng v\u1ebf c\u1ee7a (1) v\u1edbi (2), ta \u0111\u01b0\u1ee3c: <br\/> $2AM > AB - BM + AC - MC$ <br\/> $\\Rightarrow$ $AM > \\dfrac{1}{2}(AB + AC - BC)$ (*) <br\/>$\\blacktriangleright$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MA$ <br\/> X\u00e9t $\\triangle{BMA}$ v\u00e0 $\\triangle{CMD}$ c\u00f3: <br\/> $\\begin{cases} BM = MC (gt) \\\\ \\widehat{BMA} = \\widehat{CMD} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ MA = MD (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} D) \\end{cases}$ <br\/> $\\Rightarrow \\triangle{BMA} = \\triangle{CMD}$ (c.g.c) <br\/> $\\Rightarrow$ $CD = AB$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi $\\triangle{ACD}$, ta c\u00f3: <br\/> $AD < AC + CD$ <br\/> $\\Rightarrow$ $2AM < AC + AB$ <br\/> Hay $AM < \\dfrac{AC + AB}{2}$ (**) <br\/> T\u1eeb (*) v\u00e0 (**) $\\Rightarrow$ $\\dfrac{AB + AC - BC}{2} < AM < \\dfrac{AB + AC}{2}$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1887},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y \u0111i\u1ec3m $E$, tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AE = AF$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $BC + EF < BF$ ","B. $BC + EF < 2BF$ ","C. $BC + EF = 2BF$","D. $BC + EF = BF$"],"hint":"\u0110\u1ec3 xu\u1ea5t hi\u1ec7n t\u1ed5ng $BC + EF$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $EF = CK$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> \u0110\u1ec3 xu\u1ea5t hi\u1ec7n t\u1ed5ng $BC + EF$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $EF = CK$, khi \u0111\u00f3 $BC + EF = BC + CK = BK$ <br\/> Ch\u1ee9ng minh $\\widehat{E_{2}} = \\widehat{C_{2}}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $BF = FK$ b\u1eb1ng c\u00e1ch ch\u1ee9ng minh $\\triangle{BEF} = \\triangle{FCK}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho tam gi\u00e1c $BFK$, bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 c\u00f3 b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u1ea7n t\u00ecm <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K04.png' \/><\/center> <br\/> $\\blacktriangleright$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $CK = EF$. <br\/> Khi \u0111\u00f3: $BC + EF = BC + CK = BK$ (1) <br\/> $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ $\\Rightarrow$ $\\widehat{ABC} = \\widehat{C_{1}} = \\dfrac{180^{o} - \\widehat{A}}{2}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\triangle{AEF}$ c\u00e2n t\u1ea1i $A$ (v\u00ec $AE = AF$ (gt)) <br\/> $\\Rightarrow$ $\\widehat{E_{1}} = \\widehat{F_{1}} = \\dfrac{180^{o} - \\widehat{A}}{2}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $\\widehat{E_{1}} = \\widehat{C_{1}}$ n\u00ean $\\widehat{E_{2}} = \\widehat{C_{2}}$ <br\/> X\u00e9t $\\triangle{BEF}$ v\u00e0 $\\triangle{FCK}$ c\u00f3: <br\/> $\\begin{cases} EF = CK (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} \\hspace{0,2cm} K) \\\\ \\widehat{E_{2}} = \\widehat{C_{2}} (ch\u1ee9ng minh tr\u00ean) \\\\ BE = CF (AB = AC; AE = AF) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{BEF} = \\triangle{FCK}$ (c.g.c) <br\/> $\\Rightarrow$ $BF = FK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho $\\triangle{BFK}$, ta c\u00f3: <br\/> $BK < BF + FK$ (3) <br\/> T\u1eeb (1), (2), (3) suy ra: $BC + EF = BK < BF + FK = 2BF$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}]}],"id_ques":1888},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $MNP$ v\u00e0 \u0111i\u1ec3m $O$ n\u1eb1m trong tam gi\u00e1c. Ta lu\u00f4n c\u00f3 $OM + ON + OP$ l\u1edbn h\u01a1n n\u1eeda chu vi nh\u01b0ng nh\u1ecf h\u01a1n chu vi tam gi\u00e1c $MNP$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> K\u1ebb $ON$ c\u1eaft $MP$ t\u1ea1i $I$. Ch\u1ee9ng minh $ON + OM < NP + MP$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $OM + ON + OP < MN + MP + NP$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ee9ng minh $OM + ON + OP > \\dfrac{MN + MP + NP}{2}$ b\u1eb1ng c\u00e1ch \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho c\u00e1c tam gi\u00e1c $MON; MOP; NOP$ <br\/> T\u1eeb b\u01b0\u1edbc $2$ v\u00e0 b\u01b0\u1edbc $3$ suy ra \u0111i\u1ec1u c\u1ea7n t\u00ecm <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K05.png' \/><\/center> <br\/> $\\blacktriangleright$ K\u1ebb $NO$ c\u1eaft $MP$ t\u1ea1i $I$. <br\/> V\u00ec $O$ n\u1eb1m trong tam gi\u00e1c $MNP$ n\u00ean $I$ n\u1eb1m gi\u1eefa $M$ v\u00e0 $P$ hay $MP = MI + IP$ <br\/> $\\triangle{NIP}$ c\u00f3 $NI < NP + IP$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c) <br\/> $\\Rightarrow$ $NO + OI < NP + IP$ (1) <br\/> $\\triangle{MOI}$ c\u00f3 $OM < MI + OI$ (2) <br\/> C\u1ed9ng (1) v\u00e0 (2) theo t\u1eebng v\u1ebf, ta \u0111\u01b0\u1ee3c: <br\/> $NO + OI + OM < NP + IP + MI + OI$ <br\/> $\\Leftrightarrow$ $ON + OM < NP + MP$ (*) <br\/> T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: <br\/> $OM + OP < MN + NP$ (**) <br\/> $ON + OP < MN + MP$ (***) <br\/> C\u1ed9ng theo t\u1eebng v\u1ebf c\u1ee7a (*), (**), (***) ta \u0111\u01b0\u1ee3c: <br\/> $OM + ON + OP < MN + MP + NP$ (3) <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u1edbi c\u00e1c tam gi\u00e1c $MON$, $NOP$, $MOP$, ta \u0111\u01b0\u1ee3c: <br\/> $OM + ON > MN$ <br\/> $ON + OP > NP$ <br\/> $OM + OP > MP$ <br\/> C\u1ed9ng theo t\u1eebng v\u1ebf c\u1ee7a ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c tr\u00ean, ta \u0111\u01b0\u1ee3c: <br\/> $OM + ON + OP > \\dfrac{1}{2} (MN + MP + NP)$ (4) <br\/> T\u1eeb (3) v\u00e0 (4) $\\Rightarrow$ $\\dfrac{MN + MP + NP}{2} < OM + ON + OP < MN + MP + NP$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1889},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K09.png' \/><\/center> <br\/> Khi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD$ th\u00ec kho\u1ea3ng c\u00e1ch t\u1eeb $I$ \u0111\u1ebfn c\u00e1c \u0111i\u1ec3m $A$, $B$, $C$, $D$ l\u00e0 nh\u1ecf nh\u1ea5t <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> G\u1ecdi $M$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec <br\/> C\u1ea7n \u0111i ch\u1ee9ng minh $MA + MB + MC + MD \\geq AB + CD$ <br\/> \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 d\u1ea5u \"=\" x\u1ea3y ra l\u00e0 $M \\equiv I$ <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai19/lv3/img\/H7C3B19_K10.png' \/><\/center> <br\/> $\\blacktriangleright$ Theo gi\u1ea3 thi\u1ebft ta c\u00f3 $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD$, g\u1ecdi $M$ l\u00e0 \u0111i\u1ec3m b\u1ea5t k\u00ec. <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c cho c\u00e1c tam gi\u00e1c $AMB; CMD$, ta c\u00f3: <br\/> $AM + MB \\geq AB$ (d\u1ea5u \"=\" x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi $M \\in AB$) (1) <br\/> $CM + MD \\geq CD$ (d\u1ea5u \"=\" x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi $M \\in CD$) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $MA + MB + MC + MD \\geq AB + CD$ <br\/> Suy ra t\u1ed5ng $MA + MB + MC + MD$ nh\u1ecf nh\u1ea5t b\u1eb1ng $AB + CD$ khi v\u00e0 ch\u1ec9 khi $M \\equiv I$ <br\/> Hay $MA + MB + MC + MD$ nh\u1ecf nh\u1ea5t khi $M$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1890}],"lesson":{"save":0,"level":3}}