{"segment":[{"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":"<span class='basic_left'> <b> T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai. <\/b> <br\/> Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. Ta c\u00f3: ","select":["A. $\\widehat{A} = \\widehat{B} + \\widehat{C} $ ","B. $\\widehat{B} + \\widehat{C} = 90^{o} $ ","C. Hai g\u00f3c $\\widehat{B} $ v\u00e0 $\\widehat{C}$ ph\u1ee5 nhau. ","D. Hai g\u00f3c $\\widehat{B} $ v\u00e0 $\\widehat{C}$ b\u00f9 nhau. "],"hint":"Trong m\u1ed9t tam gi\u00e1c vu\u00f4ng hai g\u00f3c nh\u1ecdn ph\u1ee5 nhau. ","explain":" <span class='basic_left'> Tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A, khi \u0111\u00f3: $\\widehat{A} = 90^{o} \\\\ \\widehat{B} + \\widehat{C} = 90^{o} $ <br\/> Hai g\u00f3c $\\widehat{B} $ v\u00e0 $\\widehat{C}$ ph\u1ee5 nhau. <br\/> <br\/> <span class='basic_pink'> V\u1eady kh\u1eb3ng \u0111\u1ecbnh D l\u00e0 sai. Do \u0111\u00f3 ta ph\u1ea3i ch\u1ecdn D v\u00ec \u0111\u1ea7u b\u00e0i y\u00eau c\u1ea7u t\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai. <\/span> <\/span>","column":2}]}],"id_ques":1551},{"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 MHK vu\u00f4ng t\u1ea1i H. Ta c\u00f3: ","select":["A. $\\widehat{M} + \\widehat{K} > 90^{o} $ ","B. $\\widehat{M} + \\widehat{K} = 90^{o} $ ","C. $\\widehat{M} + \\widehat{K} < 90^{o} $ ","D. $\\widehat{M} + \\widehat{K} = 180^{o} $"],"hint":"Trong m\u1ed9t tam gi\u00e1c vu\u00f4ng hai g\u00f3c nh\u1ecdn ph\u1ee5 nhau. ","explain":" <span class='basic_left'> Tam gi\u00e1c MHK vu\u00f4ng t\u1ea1i H, khi \u0111\u00f3: $\\widehat{M} + \\widehat{K} = 90^{o} $ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span> <\/span>","column":2}]}],"id_ques":1552},{"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":" <span class='basic_left'> <b> T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai. <\/b> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-17.png' \/> <\/center> <b> T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai. <\/b> <br\/> Cho tam gi\u00e1c ABC c\u00f3 g\u00f3c $\\widehat{ACx} $ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh C c\u1ee7a tam gi\u00e1c ABC. Khi \u0111\u00f3: ","select":["A. $\\widehat{ACx} > \\widehat{A} $ ","B. $\\widehat{ACx} > \\widehat{B} $ ","C. $\\widehat{ACx} = \\widehat{A} + \\widehat{B} $ ","D. $\\widehat{ACx} + \\widehat{A} = \\widehat{B} $ "],"explain":" <span class='basic_left'> *G\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c b\u1eb1ng t\u1ed5ng hai g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3, <br\/> n\u00ean $\\widehat{ACx} = \\widehat{A} + \\widehat{B} $ <br\/> * G\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c l\u1edbn h\u01a1n m\u1ed7i g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3, <br\/> n\u00ean $\\widehat{ACx} > \\widehat{A} $ v\u00e0 $\\widehat{ACx} > \\widehat{B} $ <br\/> <br\/> <span class='basic_pink'> C\u00e2u tr\u1ea3 l\u1eddi sai l\u00e0 D. <\/span><\/span> <br\/>","column":2}]}],"id_ques":1553},{"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":" S\u1ed1 \u0111o x trong h\u00ecnh v\u1ebd sau l\u00e0: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-15.png' \/> <\/center> ","select":["A. $x = 95^{o} $ ","B. $x = 85^{o} $ ","C. $x = 80^{o} $ ","D. $x = 75^{o} $ "],"explain":" <span class='basic_left'> Ta c\u00f3 t\u1ed5ng ba g\u00f3c c\u1ee7a m\u1ed9t tam gi\u00e1c b\u1eb1ng $180^{o} $ n\u00ean: <br\/> $x = 180^{o} - 49^{o} - 36^{o} \\\\ x = 95^{o} $<br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><\/span> <br\/>","column":2}]}],"id_ques":1554},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["45"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho m\u1ed9t tam gi\u00e1c vu\u00f4ng c\u00f3 hai g\u00f3c nh\u1ecdn b\u1eb1ng nhau. <br\/> S\u1ed1 \u0111o m\u1ed7i g\u00f3c nh\u1ecdn \u0111\u00f3 l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","hint":"Trong m\u1ed9t tam gi\u00e1c vu\u00f4ng, hai g\u00f3c nh\u1ecdn ph\u1ee5 nhau.","explain":" <span class='basic_left'> Trong tam gi\u00e1c vu\u00f4ng c\u00f3 hai g\u00f3c nh\u1ecdn ph\u1ee5 nhau. <br\/> M\u00e0 theo b\u00e0i hai g\u00f3c n\u00e0y b\u1eb1ng nhau n\u00ean m\u1ed7i g\u00f3c b\u1eb1ng:<br\/> $90^{o} : 2 = 45^{o} $ <br\/> <span class='basic_pink'> \u0110\u00e1p s\u1ed1: $45^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1555},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["75"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c ABC c\u00f3 $\\widehat{B} = 70^{o}, \\widehat{C} = 40^{o} $. <br\/>Tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{A} $ c\u1eaft BC t\u1ea1i D. <br\/> S\u1ed1 \u0111o c\u1ee7a g\u00f3c $\\widehat{ADB} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$ ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-11.png' \/> <\/center> <span class='basic_left'>Trong tam gi\u00e1c ABC, ta c\u00f3: <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} \\\\ \\Leftrightarrow \\widehat{A} + 70^{o} + 40^{o} = 180^{o} \\\\ \\Leftrightarrow \\widehat{A} = 70^{o} $ <br\/> AD l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{A} $ n\u00ean $\\widehat{A_{1}} = \\dfrac{\\widehat{A}}{2} = \\dfrac{70^{o}}{2} = 35^{o} $ <br\/> Trong tam gi\u00e1c ADB, ta c\u00f3 $\\widehat{A_{1}} + \\widehat{ADB} + \\widehat{B} = 180^{o} \\\\ \\Leftrightarrow 35^{o} + \\widehat{ADB} + 70^{o} = 180^{o} \\\\ \\Leftrightarrow \\widehat{ADB} = 75^{o} $ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{ADB} = 75^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1556},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["65"],["35"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ednh c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c MNP bi\u1ebft <br\/> $\\widehat{M} = 80^{o}, \\widehat{N} - \\widehat{P} = 30^{o} $ <br\/> \u0110\u00e1p \u00e1n: $\\widehat{N} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{P} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $","explain":" <span class='basic_left'> Trong tam gi\u00e1c MNP, ta c\u00f3: <br\/> $\\widehat{M} + \\widehat{N} + \\widehat{P} = 180^{o} \\\\ \\Leftrightarrow 80^{o} + \\widehat{N} + \\widehat{P} = 180^{o} \\\\ \\Leftrightarrow \\widehat{N} + \\widehat{P} = 100^{o} (1) $ <br\/> Theo b\u00e0i: $\\widehat{N} - \\widehat{P} = 30^{o} (2) $ <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $\\widehat{N} + \\widehat{N} = 100^{o} + 30^{o} $ <br\/> $ \\Rightarrow 2\\widehat{N} = 130^{o} \\Rightarrow \\widehat{N} = 65^{o} \\\\ \\Rightarrow \\widehat{P} = 100^{o} - 65^{o} = 35^{o} $ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{N} = 65^{o} \\\\ \\widehat{P} = 35^{o} $ <\/span><\/span> <br\/>"}]}],"id_ques":1557},{"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":" <span class='basic_left'> Cho tam gi\u00e1c ABC c\u00f3 $\\widehat{B} = 30^{o}, \\widehat{C} = 50^{o} $. <br\/>Tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{A} $ c\u1eaft BC t\u1ea1i D. <br\/> S\u1ed1 \u0111o c\u1ee7a g\u00f3c $\\widehat{ADC} $ l\u00e0: ","select":["A. $80^{o} $ ","B. $120^{o} $ ","C. $100^{o} $ ","D. $110^{o} $ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-10.png' \/> <\/center> <span class='basic_left'>Trong tam gi\u00e1c $ABC$, ta c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} \\\\ \\Leftrightarrow \\widehat{A} + 30^{o} + 50^{o} = 180^{o} \\\\ \\Leftrightarrow \\widehat{A} = 100^{o} $ <br\/> $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{A} $ n\u00ean $\\widehat{A_{1}} = \\dfrac{\\widehat{A}}{2} = \\dfrac{100^{o}}{2} = 50^{o} $ <br\/> Trong tam gi\u00e1c $ADC$, ta c\u00f3 $\\widehat{A_{1}} + \\widehat{ADC} + \\widehat{C} = 180^{o} \\\\ \\Leftrightarrow 50^{o} + \\widehat{ADC} + 50^{o} = 180^{o} \\\\ \\Leftrightarrow \\widehat{ADC} = 80^{o} $ <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><\/span> <br\/>","column":4}]}],"id_ques":1558},{"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":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ v\u00e0 \u0111i\u1ec3m M n\u1eb1m trong tam gi\u00e1c \u0111\u00f3. Tia $AM$ c\u1eaft c\u1ea1nh $BC$ t\u1ea1i \u0111i\u1ec3m D. So s\u00e1nh $\\widehat{BAD} $ v\u00e0 $\\widehat{BMD} $ ","select":[" $\\widehat{BAD} < \\widehat{BMD} $ "," $\\widehat{BAD} = \\widehat{BMD} $ "," $\\widehat{BAD} > \\widehat{BMD} $ "],"hint":"G\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c l\u1edbn h\u01a1n m\u1ed1i g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3. ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-07.png' \/> <\/center> <span class='basic_left'> Tam gi\u00e1c $AMB$ c\u00f3 g\u00f3c $\\widehat{BMD} $ l\u00e0 g\u00f3c ngo\u00e0i \u0111\u1ec9nh M n\u00ean $\\widehat{BMD} > \\widehat{BAM}$ <br\/> hay $\\widehat{BMD} > \\widehat{BAD} $ <\/span> <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{BAD} < \\widehat{BMD} $ <\/span><\/span> <br\/>","column":3}]}],"id_ques":1559},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["75"],["15"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 90^{o}, \\widehat{B} = 60^{o}$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1eaft BC \u1edf D. K\u1ebb $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ (H $\\in$ BC). T\u00ednh s\u1ed1 \u0111o g\u00f3c $ADH$ v\u00e0 g\u00f3c $HAD$? <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{ADH} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{HAD} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-05.png' \/> <\/center> <span class='basic_left'> Trong tam gi\u00e1c $ABC$, ta c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} \\\\ \\Rightarrow 90^{o} + 60^{o} + \\widehat{C} = 180^{o} \\\\ \\Rightarrow \\widehat{C} = 30^{o} $ <br\/> $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{A} $ (gi\u1ea3 thi\u1ebft) <br\/> n\u00ean $\\widehat{DAC} = \\dfrac{\\widehat{A} }{2} = \\dfrac{90^{o}}{2} = 45^{o} $ <br\/> Tam gi\u00e1c ADC c\u00f3 $\\widehat{ADH} = \\widehat{DAC} + \\widehat{C} $ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> n\u00ean $\\widehat{ADH} = 45^{o} + 30^{o} = 75^{o} $ <br\/> Tam gi\u00e1c AHD vu\u00f4ng t\u1ea1i H n\u00ean $\\widehat{HAD} + \\widehat{ADH} = 90^{0} \\\\ \\Rightarrow \\widehat{HAD} = 90^{0} - 75^{o} = 15^{o} $ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{ADH} = 75^{o} \\\\ \\widehat{HAD} = 15^{o} $ <\/span><\/span> <br\/>"}]}],"id_ques":1560},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["65"],["15"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 100^{o} $ v\u00e0 $\\widehat{B} - \\widehat{C} = 50^{o}$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $B$ v\u00e0 g\u00f3c $C$? <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $\\widehat{B} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} \\\\ \\widehat{C} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","hint":"T\u1ed5ng ba g\u00f3c c\u1ee7a m\u1ed9t tam gi\u00e1c b\u1eb1ng $180^{o} $","explain":" <span class='basic_left'> Trong tam gi\u00e1c $ABC$, ta c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} \\\\ \\Rightarrow \\widehat{B} + \\widehat{C} = 180^{o} - \\widehat{A} = 180^{o} - 100^{o} = 80^{o} $ <br\/> Theo b\u00e0i c\u00f3: $\\widehat{B} - \\widehat{C} = 50^{o}$ <br\/> Do \u0111\u00f3: $\\widehat{B} + \\widehat{C} = 80^{o} \\\\ \\widehat{B} - \\widehat{C} = 50^{o} $ <br\/> $\\Rightarrow 2\\widehat{B} = 80^{o} + 50^{o} = 130^{o} \\Rightarrow \\widehat{B} = 65^{o}$ <br\/> $\\Rightarrow \\widehat{C} = 80^{o} - \\widehat{B} = 80^{o} - 65^{o} = 15^{o}$ <br\/> <br\/> <span class='basic_pink'> V\u1eady $\\widehat{B} = 65^{o} \\\\ \\widehat{C} = 15^{o} $ <\/span><\/span> <br\/>"}]}],"id_ques":1561},{"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":" <span class='basic_left'> Cho tam gi\u00e1c ABC c\u00f3 $\\widehat{B} = 80^{o}, \\widehat{C} = 30^{o} $. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1eaft BC \u1edf D. T\u00ednh s\u1ed1 \u0111o c\u1ee7a c\u00e1c g\u00f3c $\\widehat{ADC} $ v\u00e0 $\\widehat{ADB} $. ","select":["A. $\\widehat{ADC} = 110^{o}, \\widehat{ADB} = 70^{o} $ ","B. $\\widehat{ADC} = 115^{o}, \\widehat{ADB} = 65^{o} $ ","C. $\\widehat{ADC} = 120^{o}, \\widehat{ADB} = 60^{o} $ ","D. $\\widehat{ADC} = 65^{o}, \\widehat{ADB} = 115^{o} $ "],"hint":"M\u1ed7i g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c b\u1eb1ng t\u1ed5ng hai g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3. ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-04.png' \/> <\/center> <span class='basic_left'> $\\triangle ABC$ c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ n\u00ean $\\widehat{A} + 80^{o} + 30^{o} = 180^{o} $ <br\/> Suy ra $\\widehat{A} = 70^{o} $ <br\/> $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c n\u00ean $\\widehat{A_{1}} = \\widehat{A_{2}} = \\dfrac{\\widehat{A}}{2} = \\dfrac{70^{o}}{2} = 35^{o} $ <br\/> Tam gi\u00e1c $ABD$ c\u00f3 $ \\widehat{ADC} = \\widehat{B} + \\widehat{A_{1}} = 80^{o} + 35^{o} = 115^{o} $ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> M\u00e0 $\\widehat{ADB} + \\widehat{ADC} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> Suy ra $\\widehat{ADB} = 180^{o} - \\widehat{ADC} = 180^{o} - 115^{o} = 65^{o} $<br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span> <\/span>","column":2}]}],"id_ques":1562},{"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 h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-03.png' \/> <\/center> K\u1ebft qu\u1ea3 n\u00e0o sau \u0111\u00e2y l\u00e0 <b> \u0111\u00fang <\/b> : ","select":["A. $x = 97^{o}, y = 125^{o}, z = 138^{o} $ ","B. $x = 55^{o}, y = 42^{o}, z = 83^{o} $ ","C. $x + y + z = 180^{o} $ ","D. $x + y - z = 74^{o} $ "],"hint":"M\u1ed7i g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c b\u1eb1ng t\u1ed5ng hai g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3. ","explain":" Tam gi\u00e1c DEF c\u00f3 $x = \\widehat{E} + \\widehat{F} $ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> $ \\Rightarrow x = 55^{o} + 42^{o} \\\\ \\Rightarrow x = 97^{o}$ <br\/> L\u1ea1i c\u00f3 $y + 55^{o} = 180^{o} \\\\ \\Rightarrow y = 180^{o} - 55^{o} = 125^{o} \\\\ z + 42^{o} = 180^{o} \\\\ \\Rightarrow z = 180^{o} - 42^{o} = 138^{o} $ <br\/> <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 trong \u0111\u00e1p \u00e1n A l\u00e0 \u0111\u00fang nh\u1ea5t <\/span> <\/span>","column":2}]}],"id_ques":1563},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u00e2y l\u00e0 g\u00f3i 2 c\u00e2u h\u1ecfi.","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-19a.jpg' \/> <\/center> <br\/> Tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} + \\widehat{C} = \\widehat{A} $ v\u00e0 $\\widehat{C} = 2\\widehat{B}$. <br\/> Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c C c\u1eaft AB \u1edf D. <br\/> <b> C\u00e2u 1: <\/b> S\u1ed1 \u0111o g\u00f3c $\\widehat{ADC} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-19a.jpg' \/> <\/center> <span class='basic_left'>Trong tam gi\u00e1c $ABC$, ta c\u00f3 $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o} $ m\u00e0 $\\widehat{B} + \\widehat{C} = \\widehat{A} $ (gi\u1ea3 thi\u1ebft) $ \\Rightarrow 2\\widehat{A} = 180^{o} \\Rightarrow \\widehat{A} = 90^{o} $ <br\/>Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A n\u00ean $ \\widehat{B} + \\widehat{C} = 90^{o}$ m\u00e0 $\\widehat{C} = 2\\widehat{B} $ <br\/> do \u0111\u00f3 $ 3\\widehat{B} = 90^{o} \\Rightarrow \\widehat{B} = 30^{o}, \\widehat{C} = 2.\\widehat{B} = 60^{o} $ <br\/> CD l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{C} $ n\u00ean $\\widehat{ACD} = \\dfrac{\\widehat{C}}{2} = \\dfrac{60^{o}}{2} = 30^{o} $ <br\/> X\u00e9t tam gi\u00e1c vu\u00f4ng $ACD$ c\u00f3 $\\widehat{A} = 90^{o} $ <br\/> $\\Rightarrow \\widehat{ACD} + \\widehat{ADC} = 90^{o} \\Rightarrow \\widehat{ADC} = 90^{o} - 30^{o} = 60^{o} $ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p s\u1ed1: $60^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1564},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u00e2y l\u00e0 g\u00f3i 2 c\u00e2u h\u1ecfi.","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{B} + \\widehat{C} = \\widehat{A} $ v\u00e0 $\\widehat{C} = 2\\widehat{B}$. <br\/> Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c C c\u1eaft AB \u1edf D. <br\/> <b> C\u00e2u 2: <\/b> S\u1ed1 \u0111o g\u00f3c $\\widehat{BDC} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ ","hint":"","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-19a.jpg' \/> <\/center> Ta c\u00f3: $\\widehat{ADC} + \\widehat{BDC} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{BDC} = 180^{o} - \\widehat{ADC} \\\\ = 180^{o} - 60^{o} \\\\ = 120^{o} $ <\/span> <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p s\u1ed1: $120^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1565},{"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":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-20.png' \/> <\/center><br\/> Cho tam gi\u00e1c $ABC$ v\u00e0 \u0111i\u1ec3m M n\u1eb1m trong tam gi\u00e1c \u0111\u00f3. <br\/> Tia $AM$ c\u1eaft c\u1ea1nh $BC$ t\u1ea1i \u0111i\u1ec3m D. <br\/> So s\u00e1nh $\\widehat{BMD} $ v\u00e0 $\\widehat{BAD} $ ","select":[" $\\widehat{BMD} > \\widehat{BAD} $ "," $\\widehat{BMD} = \\widehat{BAD} $ "," $\\widehat{BMD} < \\widehat{BAD} $ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-20.png' \/> <\/center> <span class='basic_left'> G\u00f3c $\\widehat{BMD}$ l\u00e0 g\u00f3c ngo\u00e0i \u1edf \u0111\u1ec9nh M c\u1ee7a tam gi\u00e1c $AMB$ n\u00ean: <br\/> $\\widehat{BMD} > \\widehat{BAM}$ hay $\\widehat{BMD} > \\widehat{BAD} $ ","column":3}]}],"id_ques":1566},{"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":" <span class='basic_left'> Cho h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-22.png' \/> <\/center> S\u1ed1 \u0111o c\u1ee7a g\u00f3c $QPS$ l\u00e0: <\/span> ","select":[" A. $90^{o} $ "," B. $96^{o} $ "," C. $60^{o} $ ","D. $105^{o} $ "],"hint":"G\u00f3c ngo\u00e0i c\u1ee7a m\u1ed9t tam gi\u00e1c b\u1eb1ng t\u1ed5ng hai g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3.","explain":" <span class='basic_left'> Tam gi\u00e1c $PQR$ c\u00f3 $5x = 72^{o} + x $ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> Suy ra $5x - x = 72^{o} \\\\ \\Leftrightarrow 4x = 72^{o} \\\\ \\Leftrightarrow x = 18^{o} $ <br\/> Do \u0111\u00f3 $\\widehat{QPS} = 5x = 5.18^{o} = 90^{o} $ <br\/><br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span>","column":4}]}],"id_ques":1567},{"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":"Gi\u00e1 tr\u1ecb c\u1ee7a x \u1edf h\u00ecnh sau l\u00e0: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-23.png' \/> <\/center> ","select":[" A. $142^{o} $ "," B. $72^{o} $ "," C. $107^{o} $ ","D. $108^{o} $ "],"hint":"G\u00f3c ngo\u00e0i c\u1ee7a m\u1ed9t tam gi\u00e1c b\u1eb1ng t\u1ed5ng hai g\u00f3c trong kh\u00f4ng k\u1ec1 v\u1edbi n\u00f3.","explain":" <span class='basic_left'> Ta c\u00f3 $\\widehat{ACB} + 107^{o} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> suy ra $\\widehat{ACB} = 180^{o} - 107^{o} = 73^{o} $ <br\/> Tam gi\u00e1c ABC c\u00f3 $x = 35^{o} + 73^{o} = 108^{o} $ (g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c)<br\/><br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 D. <\/span>","column":4}]}],"id_ques":1568},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u00e2y l\u00e0 g\u00f3i 2 c\u00e2u h\u1ecfi.","temp":"fill_the_blank","correct":[[["70"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 $ \\widehat{A} = 90^{o}$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c B c\u1eaft $AC$ \u1edf E. Cho bi\u1ebft $\\widehat{C} - \\widehat{B} = 10^{o} $ <br\/> <b> C\u00e2u 1: <\/b> S\u1ed1 \u0111o g\u00f3c $\\widehat{AEB} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ <\/span> ","hint":"","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-21.png' \/> <\/center> <span class='basic_left'> Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A n\u00ean $\\widehat{B} + \\widehat{C} = 90^{o} $ (1) <br\/> M\u00e0 $\\widehat{C} - \\widehat{B} = 10^{o} $ (gi\u1ea3 thi\u1ebft) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow 2\\widehat{C} = 100^{o} \\Rightarrow \\widehat{C} = 50^{o} $ <br\/> Do \u0111\u00f3 $\\widehat{B} = 90^{o} - \\widehat{C} = 40^{o} $ <br\/> $BE$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{B}$ n\u00ean $\\widehat{B_{2}} = \\dfrac{1}{2}\\widehat{B} = \\dfrac{1}{2}.40^{o} = 20^{o} $ <br\/> G\u00f3c $\\widehat{ABE} $ l\u00e0 g\u00f3c ngo\u00e0i \u1edf \u0111\u1ec9nh E c\u1ee7a tam gi\u00e1c BEC n\u00ean: <br\/> $\\widehat{AEB} = \\widehat{C} + \\widehat{B_{2}} = 50^{o} + 20^{o} = 70^{o} $ <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p s\u1ed1: $70^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1569},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u00e2y l\u00e0 g\u00f3i 2 c\u00e2u h\u1ecfi.","temp":"fill_the_blank","correct":[[["110"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho tam gi\u00e1c ABC c\u00f3 $ \\widehat{A} = 90^{o}$. Tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c B c\u1eaft AC \u1edf E. Cho bi\u1ebft $\\widehat{C} - \\widehat{B} = 10^{o} $ <br\/> <b> C\u00e2u 2: <\/b> S\u1ed1 \u0111o g\u00f3c $\\widehat{BEC} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o} $ <\/span> ","hint":"","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai9/lv1/img\/H7B9-21.png' \/> <\/center> Ta c\u00f3: $\\widehat{AEB} = 70^{o} $ (theo c\u00e2u 1) <br\/> M\u1eb7t kh\u00e1c: $\\widehat{AEB} + \\widehat{BEC} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{BEC}$$ = 180^{o} - \\widehat{AEB} \\\\ = 180^{o} - 70^{o} \\\\ = 110^{o} $ <\/span> <br\/> <br\/> <span class='basic_pink'> \u0110\u00e1p s\u1ed1: $110^{o} $<\/span><\/span> <br\/>"}]}],"id_ques":1570}],"lesson":{"save":0,"level":1}}