{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" Cho t\u1ee9 gi\u00e1c $ABCD$ trong h\u00ecnh sau: <br\/> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_1.jpg' \/><\/center> $\\widehat{A} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ ","hint":" S\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00fd t\u1ed5ng $4$ g\u00f3c trong m\u1ed9t t\u1ee9 gi\u00e1c \u0111\u1ec3 t\u00ecm $x$, v\u00e0 suy ra s\u1ed1 \u0111o g\u00f3c $A$. ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_1.jpg' \/><\/center> <span class='basic_left'> Trong t\u1ee9 gi\u00e1c $ABCD$, ta c\u00f3: <br\/> $\\widehat{A}+\\widehat{B}+\\widehat{C}+\\widehat{D}={{360}^{o}}$ (\u0111\u1ecbnh l\u00fd) <br\/> $\\begin{align} & \\Rightarrow 2x+2x+x+x={{360}^{o}} \\\\ & \\Rightarrow 6x={{360}^{o}} \\\\ & \\Rightarrow x={{360}^{o}} :6 \\\\ & \\Rightarrow x=60^o \\\\ \\end{align}$ <br\/> Do \u0111\u00f3 $\\widehat{A}=2x=120^o$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $120$ <\/span><\/span> "}]}],"id_ques":1301},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["75"],["="]]],"list":[{"point":10,"width":40,"type_input":"","ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$, c\u00f3 $AB = AD; \\widehat{B}=90^o; \\, \\widehat{A}=60^o;\\, \\widehat{D}=135^o$ <br\/> <br\/> <b> C\u00e2u 1:<\/b> T\u00ednh s\u1ed1 \u0111o g\u00f3c $C$ v\u00e0 so s\u00e1nh $BD$ v\u00e0 $BC$. <br\/> <br\/> <b> \u0110\u00e1p \u00e1n :<\/b> <br\/> $\\widehat{C}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <br\/> $BD \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} BC$ ( \u0110i\u1ec1n $>$ ; $<$ ho\u1eb7c $=$) <\/span> ","explain":"<span class='basic_left'> V\u1ebd h\u00ecnh <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_2a.jpg' \/><\/center> X\u00e9t t\u1ee9 gi\u00e1c $ABCD$ c\u00f3: $\\widehat{A}+\\widehat{B}+\\widehat{C}+\\widehat{D}={{360}^{o}}$ (\u0111\u1ecbnh l\u00ed) <br\/> $\\Rightarrow \\widehat{C}={{360}^{o}}-\\left( {{60}^{o}}+{{90}^{o}}+{{135}^{o}} \\right)={{75}^{o}}$ <br\/> X\u00e9t $\\Delta ABD$ c\u00f3: $AB = AD$ (gi\u1ea3 thi\u1ebft); $\\widehat{A}={{60}^{o}}$ <br\/>$\\Rightarrow \\Delta ABD$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u. (t\u00ednh ch\u1ea5t)$\\Rightarrow \\widehat{{{D}_{2}}}={{60}^{o}}$<br\/> M\u00e0 $\\widehat{ADC}=\\widehat{{{D}_{1}}}+\\widehat{{{D}_{2}}}$ <br\/> $\\Rightarrow \\widehat{{{D}_{1}}}={{135}^{o}}-{{60}^{o}}={{75}^{o}}$<br\/> X\u00e9t $\\Delta BDC$ c\u00f3: $\\widehat{C}=\\widehat{{{D}_{1}}}={{75}^{o}}$ <br\/> $\\Rightarrow \\Delta BDC$ c\u00e2n t\u1ea1i $B$ (t\u00ednh ch\u1ea5t) <br\/>$\\Rightarrow BD=BC$ (t\u00ednh ch\u1ea5t) <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $\\widehat{C}=75^o$ v\u00e0 $BD=BC$ <\/span><\/span> "}]}],"id_ques":1302},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["30"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$, c\u00f3 $AB = AD; \\widehat{B}=90^o; \\, \\widehat{A}=60^o;\\, \\widehat{D}=135^o$ <br\/> <br\/> <b> C\u00e2u 2:<\/b> T\u1eeb $A$ k\u1ebb $AE \\bot CD$ t\u1ea1i $E$. Trong tam gi\u00e1c $ACE$ th\u00ec $\\widehat{ACE}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span>","hint":" T\u00ednh s\u1ed1 \u0111o g\u00f3c $BCA$, sau \u0111\u00f3 suy ra s\u1ed1 \u0111o g\u00f3c $ACE$.","explain":"<span class='basic_left'> V\u1ebd h\u00ecnh <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_2bc.jpg' \/><\/center> Ta c\u00f3 $AB = BD$ (t\u00ednh ch\u1ea5t) ; $BC= BD$ (ch\u1ee9ng minh \u1edf c\u00e2u 1)<br\/> $\\Rightarrow AB = BC \\Rightarrow \\Delta ABC$vu\u00f4ng c\u00e2n (\u0111\u1ecbnh ngh\u0129a)<br\/> $\\Rightarrow \\widehat{BCA}={{45}^{o}}$ <br\/> M\u1eb7t kh\u00e1c : $\\widehat{C}=\\widehat{BCA}+\\widehat{ACD}$<br\/> $\\Rightarrow \\widehat{ACD}={{75}^{o}}-{{45}^{o}}={{30}^{o}}$ <br\/> Do \u0111\u00f3 $\\widehat{ACE}=30^o$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $30$ <\/span><\/span> "}]}],"id_ques":1303},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$, c\u00f3 $AB = AD; \\widehat{B}=90^o; \\, \\widehat{A}=60^o;\\, \\widehat{D}=135^o$ <br\/> <br\/> <b> C\u00e2u 3:<\/b> T\u1eeb $A$ k\u1ebb $AE \\bot CD$ t\u1ea1i $E$. <br\/> Khi \u0111\u00f3: $\\widehat{EAC}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span>","hint":" X\u00e9t trong tam gi\u00e1c $AEC$, t\u00ednh s\u1ed1 \u0111o g\u00f3c $EAC$.","explain":"<span class='basic_left'> V\u1ebd h\u00ecnh <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_2bc.jpg' \/><\/center> Theo c\u00e2u 2, ta t\u00ednh \u0111\u01b0\u1ee3c: $\\widehat{ACE}=30^o$ <br\/> X\u00e9t $\\Delta AEC$ c\u00f3: $\\widehat{E}={{90}^{o}}$<br\/>$\\Rightarrow \\widehat{EAC}+\\widehat{ACE}={{90}^{o}}$ (hai g\u00f3c nh\u1ecdn ph\u1ee5 nhau)<br\/> $\\Rightarrow \\widehat{EAC}={{90}^{o}}-{{30}^{o}}={{60}^{o}}$<br\/> <span class='basic_pink'> Do \u0111\u00f3 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $60$ <\/span><\/span> "}]}],"id_ques":1304},{"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":" <span class='basic_left'> Cho m\u1ed9t t\u1ee9 gi\u00e1c c\u00f3 chu vi l\u00e0 $80\\, cm$, \u0111\u1ed9 d\u00e0i m\u1ed9t trong c\u00e1c c\u1ea1nh l\u1edbn h\u01a1n \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh c\u00f2n l\u1ea1i l\u00e0 $3\\, cm; 4\\, cm; 5\\, cm$. \u0110\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh c\u1ee7a t\u1ee9 gi\u00e1c \u0111\u00f3 l\u00e0: <\/span>","select":["A. $23 cm;\\, 20 cm;\\, 19 cm;\\, 18 cm$ ","B. $23 cm;\\, 21 cm;\\, 19 cm;\\, 17 cm$ ","C. $22 cm;\\, 20 cm;\\, 19 cm;\\, 18 cm$ "],"hint":" G\u1ecdi c\u1ea1nh l\u1edbn nh\u1ea5t c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 $x$, c\u00e1c c\u1ea1nh c\u00f2n l\u1ea1i l\u00e0 $x - 3; x - 4; x - 5$. ","explain":" <span class='basic_left'> G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh l\u1edbn nh\u1ea5t l\u00e0 $x\\, (cm)$ ($x > 5; x \\in \\mathbb {R}$)<br\/> \u0110\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh c\u00f2n l\u1ea1i l\u00e0 $x-3$; $x-4$ v\u00e0 $x-5$ <br\/> Chu vi c\u1ee7a t\u1ee9 gi\u00e1c \u0111\u00f3 l\u00e0 $80 \\,cm$ n\u00ean: <br\/> $x+x-3+x-4+x-5=80$ <br\/> $\\Rightarrow 4x-12=80$ <br\/> $\\Rightarrow 4x=92 $ <br\/> $\\Rightarrow x= 23$ (th\u1ecfa m\u00e3n) <br\/> Do \u0111\u00f3 c\u00e1c c\u1ea1nh c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0: $23 cm;\\, 20 cm;\\, 19 cm$ v\u00e0 $18 cm$. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":1}]}],"id_ques":1305},{"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":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$ c\u00f3 $\\widehat{A}=150^o$. T\u1ed5ng s\u1ed1 \u0111o $3$ g\u00f3c ngo\u00e0i c\u00f2n l\u1ea1i c\u1ee7a t\u1ee9 gi\u00e1c l\u00e0: <\/span>","select":["A. $220^o$ ","B. $330^o$ ","C. $293^o$ ","D. $310^o$ "],"hint":" T\u00ednh s\u1ed1 \u0111o g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh A. <br\/> <b> L\u01b0u \u00fd: <\/b> T\u1ed5ng s\u1ed1 \u0111o c\u00e1c g\u00f3c ngo\u00e0i c\u1ee7a t\u1ee9 gi\u00e1c l\u00e0 $360^o$ ","explain":" <span class='basic_left'> S\u1ed1 \u0111o g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $A$ c\u1ee7a t\u1ee9 gi\u00e1c l\u00e0: <br\/> $180^o-150^o=30^o$ <br\/> M\u00e0 t\u1ed5ng c\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i c\u00e1c \u0111\u1ec9nh t\u1ee9 gi\u00e1c b\u1eb1ng $360^o$. <br\/> Do \u0111\u00f3 t\u1ed5ng s\u1ed1 \u0111o $3$ g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $B, C$ v\u00e0 $D$ l\u00e0: <br\/> $360^o-30^o=330^o$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 $B$. <\/span><\/span> ","column":4}]}],"id_ques":1306},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$, v\u1ebd $AD \\cap BC$ t\u1ea1i $E$ v\u00e0 $BA\\cap CD$ t\u1ea1i $F$. Hai tia ph\u00e2n gi\u00e1c c\u00e1c g\u00f3c $AEB$ v\u00e0 $BFC$ c\u1eaft nhau t\u1ea1i $K$. <br\/> <br\/> <b> C\u00e2u 1:<\/b> Ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c $\\widehat{EKF}=\\dfrac{\\widehat {B}+\\widehat{D}}{2}$ <\/span> ","select":["\u0110\u00fang","Sai"],"hint":" T\u00ednh g\u00f3c $EKF$ theo c\u00e1c g\u00f3c $EFK$ v\u00e0 $FEK$. <br\/> C\u00e1c g\u00f3c $EFK$ v\u00e0 $FEK$ \u0111\u01b0\u1ee3c chia th\u00e0nh c\u00e1c g\u00f3c nh\u1ecf \u0111\u1ec3 t\u00ednh theo c\u00e1c g\u00f3c l\u1edbn $A, B, C$ v\u00e0 $D$ trong t\u1ee9 gi\u00e1c $ABCD$. ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_3.jpg' \/><\/center> X\u00e9t trong tam gi\u00e1c $ABE$ c\u00f3: <br\/> $2{{\\widehat{E}}_{1}}={{180}^{o}}-\\left( \\widehat{A}+\\widehat{B} \\right)$ <br\/> X\u00e9t trong tam gi\u00e1c $BCF$ c\u00f3: <br\/> $2{{\\widehat{F}}_{1}}={{180}^{o}}-\\left( \\widehat{C}+\\widehat{B} \\right)$<br\/> X\u00e9t trong tam gi\u00e1c $EFD$: <br\/> ${{\\widehat{E}}_{2}}+{{\\widehat{F}}_{2}}={{180}^{o}}-\\widehat{D}$ <br\/> M\u1eb7t kh\u00e1c , trong tam gi\u00e1c $EFK$: <br\/> $\\begin{align} & \\widehat{EKF}={{180}^{o}}-\\left( {{\\widehat{E}}_{1}}+{{\\widehat{E}}_{2}}+{{\\widehat{F}}_{1}}+{{\\widehat{F}}_{2}} \\right) \\\\ & ={{180}^{o}}-\\left[ \\left( {{\\widehat{E}}_{1}}+{{\\widehat{F}}_{1}} \\right)+\\left( {{\\widehat{E}}_{2}}+{{\\widehat{F}}_{2}} \\right) \\right] \\\\ & ={{180}^{o}}-\\left[ \\dfrac{{{180}^{o}}-\\left( \\widehat{A}+\\widehat{B} \\right)}{2}+\\dfrac{{{180}^{o}}-\\left( \\widehat{C}+\\widehat{B} \\right)}{2}+{{180}^{o}}-\\widehat{D} \\right] \\\\ & =- \\dfrac{{{180}^{o}}-\\left( \\widehat{A}+\\widehat{B} \\right)}{2}-\\dfrac{{{180}^{o}}-\\left( \\widehat{C}+\\widehat{B} \\right)}{2}+\\widehat{D} \\\\ & =-90^{o}+\\dfrac{\\widehat{A}+\\widehat{B}}{2}-90^{o}+\\dfrac{\\widehat{C}+\\widehat{B}}{2}+\\widehat{D} \\\\ &=-180^{o}+\\dfrac{\\widehat{A}}{2}+\\widehat{B}+\\dfrac{\\widehat{C}}{2}+\\widehat{D}\\\\ & =-{{180}^{o}}+\\dfrac{\\widehat{A}+\\widehat{B}+\\widehat{C}+\\widehat{D}+\\widehat{B}+\\widehat{D}}{2} \\\\ & =-{{180}^{8}}+\\dfrac{{{360}^{o}}+\\widehat{B}+\\widehat{D}}{2} \\\\ & =\\dfrac{\\widehat{B}+\\widehat{D}}{2} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0: \u0110\u00fang.<\/span>","column":2}]}],"id_ques":1307},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":40,"type_input":"","ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$, v\u1ebd $AD \\cap BC$ t\u1ea1i $E$ v\u00e0 $BA\\cap CD $ t\u1ea1i $F$. Hai tia ph\u00e2n gi\u00e1c c\u00e1c g\u00f3c $AEB$ v\u00e0 $BFC$ c\u1eaft nhau t\u1ea1i $K$. <br\/> <br\/> <b> C\u00e2u 2:<\/b> N\u1ebfu $\\widehat{A}+\\widehat{C}=180^o$ th\u00ec s\u1ed1 \u0111o $\\widehat{EKF}$ b\u1eb1ng bao nhi\u00eau? <br\/> <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> <br\/> $\\widehat{EKF}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$ <\/span>","explain":"<span class='basic_left'> V\u1ebd h\u00ecnh <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_3.jpg' \/><\/center> Theo b\u00e0i: <br\/> $\\begin{align} & \\widehat{A}+\\widehat{C}={{180}^{o}} \\\\ & \\Rightarrow \\widehat{B}+\\widehat{D}={{180}^{o}} \\\\ \\end{align}$ <br\/> M\u00e0 theo c\u00e2u 1, ta t\u00ednh \u0111\u01b0\u1ee3c: $\\widehat{EKF}=\\dfrac{\\widehat{B}+\\widehat{D}}{2}$ <br\/> Do \u0111\u00f3: <br\/> $\\widehat{EKF}=\\dfrac{{{180}^{o}}}{2}={{90}^{o}}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $90$.<\/span><\/span> "}]}],"id_ques":1308},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$, v\u1ebd $AD \\cap BC$ t\u1ea1i $E$ v\u00e0 $BA\\cap CD $ t\u1ea1i $F$. Hai tia ph\u00e2n gi\u00e1c c\u00e1c g\u00f3c $AEB$ v\u00e0 $BFC$ c\u1eaft nhau t\u1ea1i $K$. <br\/> <br\/> <b> C\u00e2u 3:<\/b> N\u1ebfu $\\widehat{A}+\\widehat{C}=180^o$ th\u00ec hai tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c $AFD$ v\u00e0 $CED$ vu\u00f4ng g\u00f3c v\u1edbi nhau. <\/span> ","select":["\u0110\u00fang","Sai"],"hint":" T\u1eeb c\u00e2u 2 \u0111\u00e3 l\u00e0m tr\u00ean ta suy ra. ","explain":"<span class='basic_left'> V\u1ebd h\u00ecnh <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_3.jpg' \/><\/center> Theo c\u00e2u 2, ta t\u00ednh \u0111\u01b0\u1ee3c: $\\widehat{EKF}=90^o$ <br\/> $\\Rightarrow FK \\bot EK$ <br\/> <i>K\u1ebft lu\u1eadn: <\/i> Hai tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c $AFD$ v\u00e0 $CED$ vu\u00f4ng g\u00f3c v\u1edbi nhau. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":1309},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Trong m\u1ed9t t\u1ee9 gi\u00e1c, m\u1ed7i \u0111\u01b0\u1eddng ch\u00e9o nh\u1ecf h\u01a1n n\u1eeda chu vi t\u1ee9 gi\u00e1c. <\/span> ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <i>\u0110\u1ed1i v\u1edbi \u0111\u01b0\u1eddng ch\u00e9o $AC$:<\/i> <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c trong tam gi\u00e1c $ABC$ v\u00e0 $ADC$. <br\/> T\u1eeb \u0111\u00f3 suy ra \u0111i\u1ec1u c\u1ea7n ch\u1ee9ng minh. <br\/> T\u01b0\u01a1ng t\u1ef1 \u0111\u1ed1i v\u1edbi \u0111\u01b0\u1eddng ch\u00e9o $BD$ <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai1/lv3/img\/H811_K2_4.jpg' \/><\/center> X\u00e9t t\u1ee9 gi\u00e1c $ABCD$ c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o $AC$ v\u00e0 $BD$. <br\/> Ta ch\u1ee9ng minh v\u1edbi \u0111\u01b0\u1eddng ch\u00e9o $AC$ (\u0111\u01b0\u1eddng ch\u00e9o $BD$ t\u01b0\u01a1ng t\u1ef1) <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c trong hai tam gi\u00e1c $ABC$ v\u00e0 $ACD$, ta c\u00f3: <br\/> $ AC < AB + BC$ <br\/> $AC < AD + DC$ <br\/> C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf hai b\u1ea5t \u0111\u1eb3ng th\u1ee9c tr\u00ean: <br\/> $\\Rightarrow 2AC < AB + BC +AD+DC$ <br\/> $\\Rightarrow AC < \\dfrac{AB+BC+AD+DC}{2}$ <br\/>T\u01b0\u01a1ng t\u1ef1 \u0111\u1ed1i v\u1edbi \u0111\u01b0\u1eddng ch\u00e9o $BD$. <br\/> <i> K\u1ebft lu\u1eadn: <\/i> Trong m\u1ed9t t\u1ee9 gi\u00e1c, m\u1ed7i \u0111\u01b0\u1eddng ch\u00e9o nh\u1ecf h\u01a1n n\u1eeda chu vi t\u1ee9 gi\u00e1c. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":1310}],"lesson":{"save":0,"level":3}}