{"segment":[{"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":[[["28"],["12"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Qua \u0111i\u1ec3m A tr\u00ean m\u1eb7t ph\u1eb3ng v\u1ebd b\u1ed1n \u0111\u01b0\u1eddng th\u1eb3ng ph\u00e2n bi\u1ec7t. <br\/> S\u1ed1 g\u00f3c \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} (g\u00f3c)$ <br\/> S\u1ed1 c\u1eb7p g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh kh\u00e1c g\u00f3c b\u1eb9t l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} (g\u00f3c)$","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh l\u00e0 hai g\u00f3c m\u00e0 m\u1ed7i c\u1ea1nh c\u1ee7a g\u00f3c n\u00e0y l\u00e0 tia \u0111\u1ed1i m\u1ed9t c\u1ea1nh c\u1ee7a g\u00f3c kia. <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> 4 \u0111\u01b0\u1eddng th\u1eb3ng qua \u0111i\u1ec3m A t\u1ea1o th\u00e0nh 8 tia, trong \u0111\u00f3 m\u1ed7i tia t\u1ea1o v\u1edbi 7 tia c\u00f2n l\u1ea1i 7 g\u00f3c, <br\/> v\u1eady $8$ tia t\u1ea1o th\u00e0nh: $8.7 = 56 $(g\u00f3c) <br\/> Tuy nhi\u00ean m\u1ed7i g\u00f3c \u0111\u01b0\u1ee3c t\u00ednh 2 l\u1ea7n, n\u00ean s\u1ed1 g\u00f3c \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh l\u00e0: $56 : 2 = 28$ (g\u00f3c) <br\/> Trong $28$ g\u00f3c t\u1ea1o th\u00e0nh c\u00f3 $4$ g\u00f3c b\u1eb9t, n\u00ean s\u1ed1 g\u00f3c kh\u00e1c g\u00f3c b\u1eb9t l\u00e0: $ 28 - 4 = 24$ (g\u00f3c) <br\/>M\u1ed7i g\u00f3c trong 24 g\u00f3c n\u00e0y \u0111\u1ec1u c\u00f3 m\u1ed9t g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh v\u1edbi n\u00f3, <br\/> V\u1eady s\u1ed1 c\u1eb7p g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh kh\u00e1c g\u00f3c b\u1eb9t l\u00e0. $24 : 2 = 12$ (c\u1eb7p g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <\/span> <br\/><br\/><span class='basic_pink'> \u0110\u00e1p s\u1ed1 l\u00e0 28, 12 <\/span><\/span> <span class='basic_left'><b> L\u01b0u \u00fd <\/b> <br\/> Ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh s\u1ed1 g\u00f3c \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh t\u1eeb n \u0111\u01b0\u1eddng th\u1eb3ng l\u00e0: n(2n-1) <br\/> S\u1ed1 c\u1eb7p g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh (kh\u00f4ng k\u1ec3 g\u00f3c b\u1eb9t) l\u00e0: n(n - 1) <\/span>"}]}],"id_ques":1351},{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","t","t"]],"list":[{"point":10,"image":"","ques":"<span class='basic_left'>Cho \u0111i\u1ec3m O n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng aa'. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng c\u00f3 b\u1edd l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng aa' v\u1ebd tia Ob sao cho $\\widehat{aOb} = 45^{o}; $ c\u00f2n tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng kia v\u1ebd tia Oc sao cho $\\widehat{aOc} = 90^{o}. $ G\u1ecdi Ob' l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{a'Oc}$. C\u00e1c k\u1ebft lu\u1eadn sau \u0111\u00fang hay sai? <\/span>","col_name":["K\u1ebft lu\u1eadn","\u0110\u00fang","Sai"],"arr_ques":["Oa v\u00e0 Oa' l\u00e0 hai tia \u0111\u1ed1i nhau","$\\widehat{a'Ob'} = 60^{o} $","Ob v\u00e0 Ob' l\u00e0 hai tia \u0111\u1ed1i nhau","$\\widehat{a'Ob}$ v\u00e0 $\\widehat{aOb'}$ l\u00e0 hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh"],"hint":"","explain":["\u0110\u00fang. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-38.png' \/><\/center>","<br\/> Sai. V\u00ec Ob' l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{a'Oc}$ (gi\u1ea3i thi\u1ebft)<br\/>Suy ra $\\widehat{a'Ob'} = \\widehat{b'Oc} = \\frac{1}{2} . \\widehat{a'Oc} = \\frac{1}{2} . 90^{o} = 45^{o}$ ","<br\/> \u0110\u00fang. V\u00ec:<br\/>Ta c\u00f3: $\\widehat{aOc} + \\widehat{a'Oc} = \\widehat{aOa'} = 18^{o}$<br\/>Suy ra $\\widehat{a'Oc} = 180^{o} - \\widehat{aOc} = 180^{o} - 90^{o} = 90^{o}$<br\/>L\u1ea1i c\u00f3: Ob' l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{a'Oc}$ (gi\u1ea3i thi\u1ebft)<br\/>Suy ra $\\widehat{a'Ob'} = \\widehat{b'Oc} = \\frac{1}{2} . \\widehat{a'Oc} = \\frac{1}{2} . 90^{o} = 45^{o}$<br\/>$\\widehat{b'Ob} = \\widehat{b'Oc} + \\widehat{aOc} + \\widehat{aOb} = 45^{o} + 90^{o} + 45^{o} = 180^{o}$<br\/>Suy ra Ob v\u00e0 Ob' l\u00e0 hai tia \u0111\u1ed1i nhau.","<br\/> \u0110\u00fang. V\u00ec Oa v\u00e0 Oa' l\u00e0 hai tia \u0111\u1ed1i nhau v\u00e0 Ob v\u00e0 Ob' l\u00e0 hai tia \u0111\u1ed1i nhau"]}]}],"id_ques":1352},{"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":[[["80"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho hai \u0111\u01b0\u1eddng th\u1eb3ng xx' v\u00e0 yy' c\u1eaft nhau t\u1ea1i O sao cho t\u1ec9 s\u1ed1 s\u1ed1 \u0111o c\u1ee7a g\u00f3c $\\widehat{xOy} $ v\u00e0 $\\widehat{xOy'} $ l\u00e0 5:4. <br\/> S\u1ed1 \u0111o g\u00f3c $\\widehat{x'Oy} $ l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh th\u00ec b\u1eb1ng nhau. <br\/> Hai g\u00f3c k\u1ec1 b\u00f9 c\u00f3 t\u1ed5ng b\u1eb1ng $180^{o}$ <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-37.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $\\dfrac{\\widehat{xOy}}{\\widehat{xOy'}} = \\dfrac{5}{4} \\Rightarrow \\widehat{xOy} = \\dfrac{5}{4}\\widehat{xOy'} $ <br\/> M\u00e0 $\\widehat{xOy} + \\widehat{xOy'} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> N\u00ean $\\dfrac{5}{4}\\widehat{xOy'} + \\widehat{xOy'} = 180^{o} \\\\ \\dfrac{9}{4}\\widehat{xOy'} = 180^{o} \\\\ \\widehat{xOy'} = 80^{o} $ <br\/> V\u00ec $\\widehat{xOy'} = \\widehat{x'Oy}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) n\u00ean $\\widehat{x'Oy} = 80^{o} $<\/span> <br\/><br\/><span class='basic_pink'> \u0110\u00e1p s\u1ed1 l\u00e0: $ 80^{o}. $ <\/span><\/span> "}]}],"id_ques":1353},{"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":[[["135"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11_47.png' \/><\/center>Cho h\u00ecnh v\u1ebd nh\u01b0 tr\u00ean, bi\u1ebft $\\widehat{O_{1}} = 3\\widehat{O_{2}}$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{O_{4}}$? <br\/> \u0110\u00e1p s\u1ed1: $\\widehat{O_{4}} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$ <br\/> ","hint":"T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{O_{2}}$ r\u1ed3i suy ra $\\widehat{O_{4}}$.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh th\u00ec b\u1eb1ng nhau. <br\/> Hai g\u00f3c k\u1ec1 b\u00f9 c\u00f3 t\u1ed5ng b\u1eb1ng $180^{o}$ <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11_47.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $\\widehat{O_{1}} + \\widehat{O_{2}} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> M\u00e0 $\\widehat{O_{1}} = 3\\widehat{O_{2}}$ (gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3 $ 3\\widehat{O_{2}} + \\widehat{O_{2}} = 180^{o} \\\\ \\Rightarrow 4\\widehat{O_{2}} = 180^{o} \\\\ \\Rightarrow \\widehat{O_{2}} = 45^{o} $ <br\/> L\u1ea1i c\u00f3 $\\widehat{O_{4}} + \\widehat{O_{2}} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9)<br\/>Suy ra $\\widehat{O_{4}} = 180^{o} - 45^{o} = 135^{o}$<\/span> <br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: <span class='basic_pink'> 135<\/span><\/span> "}]}],"id_ques":1354},{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Hai tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh l\u00e0 hai tia \u0111\u1ed1i nhau? ","select":[" \u0110\u00fang "," Sai "],"hint":"","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-34.png' \/><\/center> <br\/> $\\widehat{xOy}$ v\u00e0 $\\widehat{x'Oy'}$ l\u00e0 hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh. <br\/. V\u1ebd tia Ot l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{x'Oy}$ <br\/> Ta c\u00f3 Ot v\u00e0 Oz l\u00e0 hai tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c k\u1ec1 b\u00f9 $\\widehat{x'Oy}$ v\u00e0 $\\widehat{xOy}$, do \u0111\u00f3 $\\widehat{zOt} = 90^{o} $ <br\/> M\u1eb7t kh\u00e1c Ot v\u00e0 Oz' l\u00e0 hai tia ph\u00e2n gi\u00e1c c\u1ee7a hai g\u00f3c k\u1ec1 b\u00f9 $\\widehat{x'Oy}$ v\u00e0 $\\widehat{x'Oy'}$, do \u0111\u00f3 $\\widehat{z'Ot} = 90^{o} $ <br\/> N\u00ean $\\widehat{zOt} + \\widehat{z'Ot} = 90^{o} + 90^{o} = 180^{o} $ <br\/> M\u00e0 hai tia Oz, Oz' kh\u00f4ng tr\u00f9ng nhau do \u0111\u00f3 Oz v\u00e0 Oz' l\u00e0 hai tia \u0111\u1ed1i nhau. <br\/><br\/><span class='basic_pink'> \u0110\u00e1p \u00e1n l\u00e0 \u0110\u00fang. <\/span><\/span> ","column":4}]}],"id_ques":1355},{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho g\u00f3c b\u1eb9t AOB. Tr\u00ean c\u00f9ng m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd AB, ta v\u1ebd hai tia OC v\u00e0 OD sao cho $\\widehat{AOC} = \\widehat{BOD} = 150^{o} $. G\u1ecdi tia OE l\u00e0 tia \u0111\u1ed1i c\u1ee7a tia OD <br\/> Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang? <\/span> ","select":["A. Tia OC l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{DOB} $ ","B. $\\widehat{AOD}$ v\u00e0 $\\widehat{BOC}$ l\u00e0 hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh. ","C. Tia OB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{EOC} $ ","D. $\\widehat{COD}$ v\u00e0 $\\widehat{AOE}$ l\u00e0 hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh. ","E. Tia OD l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{AOC} $ "],"hint":"V\u1ebd h\u00ecnh r\u1ed3i t\u00ecm ph\u00e1t bi\u1ec3u \u0111\u00fang.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh l\u00e0 hai g\u00f3c m\u00e0 m\u1ed7i c\u1ea1nh c\u1ee7a g\u00f3c n\u00e0y l\u00e0 tia \u0111\u1ed1i m\u1ed9t c\u1ea1nh c\u1ee7a g\u00f3c kia. <\/span> <br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-30.png' \/><\/center>Ta c\u00f3: $\\widehat{AOC} + \\widehat{BOC} = 180^{o}$ (hai g\u00f3c k\u1ec1 b\u00f9)<br\/>$\\Rightarrow$ $\\widehat{BOC} = 180^{o} - \\widehat{AOC} = 180^{o} - 150^{o} = 30^{o}$ (1)<br\/>L\u1ea1i c\u00f3: $\\widehat{BOC} + \\widehat{COD} = \\widehat{BOD}$<br\/>$\\Rightarrow$ $\\widehat{COD} = \\widehat{BOD} - \\widehat{BOC} = 150^{o} - 30^{o} = 120^{o}$ ( 2)<br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{BOC} < \\widehat{COD}$<br\/>$\\Rightarrow$ <b>Tia OC kh\u00f4ng ph\u1ea3i l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{DOB} $<\/b> <span class='basic_pink'>(\u0111\u00e1p \u00e1n A sai)<\/span><br\/>L\u1ea1i c\u00f3: OE l\u00e0 tia \u0111\u1ed1i c\u1ee7a tia OD (gi\u1ea3 thi\u1ebft)<br\/> $\\Rightarrow$ $\\widehat{EOB} + \\widehat{BOD} = 180^{o}$ (hai g\u00f3c k\u1ec1 b\u00f9)<br\/>$\\Rightarrow$ $\\widehat{EOB} = 180^{o} - \\widehat{BOD} = 180^{o} - 150^{o} = 30^{o}$ (3)<br\/>T\u1eeb (1) v\u00e0 (3) $\\Rightarrow$ $\\widehat{EOB} = \\widehat{EOB}$<br\/>M\u00e0 tia OB n\u1eb1m gi\u1eefa hai tia OE v\u00e0 OC<br\/>$\\Rightarrow$ <b>Tia OB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{EOC} $ <\/b> <span class='basic_pink'>(\u0111\u00e1p \u00e1n C \u0111\u00fang)<br\/>$\\widehat{AOD} + \\widehat{BOD} = 180^{o}$ (hai g\u00f3c k\u1ec1 b\u00f9)<br\/>$\\Rightarrow$ $\\widehat{AOD} = 180^{o} - \\widehat{BOD} = 180^{o} - 150^{o} = 30^{o}$ (4)<br\/>T\u1eeb (2) v\u00e0 (4) $\\Rightarrow$ $\\widehat{AOD} < \\widehat{COD}$ <br\/>$\\Rightarrow$ Tia OD kh\u00f4ng ph\u1ea3i l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{AOC} $ <span class='basic_pink'>(\u0111\u00e1p \u00e1n D sai)<\/span><br\/>$\\widehat{AOD}$ v\u00e0 $\\widehat{BOC}$ kh\u00f4ng ph\u1ea3i l\u00e0 hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh v\u00ec OD v\u00e0 OC kh\u00f4ng ph\u1ea3i l\u00e0 hai tia \u0111\u1ed1i nhau <span class='basic_pink'>(\u0111\u00e1p \u00e1n B sai)<\/span><\/span><br\/><\/span> <br\/><\/span> ","column":1}]}],"id_ques":1356},{"time":14,"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'>Hai \u0111\u01b0\u1eddng th\u1eb3ng xy v\u00e0 mn c\u1eaft nhau t\u1ea1i O. V\u1ebd tia ph\u00e2n gi\u00e1c Ot c\u1ee7a g\u00f3c xOm. Bi\u1ebft $ \\widehat{yOt} = 125^{o}. $<br\/> T\u00ednh s\u1ed1 \u0111o c\u00e1c g\u00f3c $\\widehat{xOm}$ v\u00e0 $\\widehat{yOn}$? <\/span> ","select":[" A. $110^{o}, 125^{o} $ "," B. $110^{o}, 110^{o} $ "," C. $55^{o}, 55^{o} $"," D. $110^{o}, 80^{o} $"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh th\u00ec b\u1eb1ng nhau. <br\/> Hai g\u00f3c k\u1ec1 b\u00f9 c\u00f3 t\u1ed5ng b\u1eb1ng $180^{o}$ <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-22.png' \/><\/center> Ta c\u00f3: $\\widehat{xOt} + \\widehat{yOt} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> Suy ra: $\\widehat{xOt}$$ = 180^{o} - \\widehat{yOt} \\\\ = 180^{o} - 125^{o} \\\\ = 55^{o} $ <br\/> Do Ot l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat{xOm}$ n\u00ean $\\widehat{xOm} = 2\\widehat{xOt} = 2. 55^{o} = 110^{o} $ <br\/> M\u00e0 $\\widehat{xOm}$ v\u00e0 $\\widehat{yOn}$ l\u00e0 hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh n\u00ean: $\\widehat{xOm} = \\widehat{yOn} = 110^{o} $ <\/span> <br\/><br\/><span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":1357},{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'>Hai \u0111\u01b0\u1eddng th\u1eb3ng AB v\u00e0 CD c\u1eaft nhau t\u1ea1i O t\u1ea1o th\u00e0nh 4 g\u00f3c kh\u00f4ng k\u1ec3 g\u00f3c b\u1eb9t. Bi\u1ebft t\u1ed5ng c\u1ee7a 3 trong 4 g\u00f3c n\u00e0y b\u1eb1ng $250^{o}.$ T\u00ednh s\u1ed1 \u0111o 4 g\u00f3c \u0111\u00f3? <\/span><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-21.png' \/><\/center> ","select":[" A. $70^{o}, 90^{o}, 70^{o}, 90^{o} $ "," B. $140^{o}, 40^{o}, 140^{o}, 40^{o} $ "," C. $70^{o}, 110^{o}, 70^{o}, 110^{o} $"," D. $130^{o}, 50^{o}, 130^{o}, 50^{o} $"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh th\u00ec b\u1eb1ng nhau. <br\/> Hai g\u00f3c k\u1ec1 b\u00f9 c\u00f3 t\u1ed5ng b\u1eb1ng $180^{o}$ <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> Gi\u1ea3 s\u1eed $\\widehat{O_{1}} + \\widehat{O_{2}} + \\widehat{O_{3}} = 250^{o} $ <br\/> V\u00ec $\\widehat{O_{1}} + \\widehat{O_{2}} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{O_{3}} = 250^{o} - 180^{o} = 70^{0} $ <br\/> V\u00ec $\\widehat{O_{2}} + \\widehat{O_{3}} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{O_{2}} = 180^{o} - \\widehat{O_{3}} \\\\ = 180^{o} - 70^{o} = 110^{0} $ <br\/> V\u1eady: $\\widehat{O_{1}} = \\widehat{O_{3}} = 70^{o}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $\\quad \\widehat{O_{2}} = \\widehat{O_{4}} = 110^{0} $ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <\/span> <br\/><br\/><span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><\/span> ","column":2}]}],"id_ques":1358},{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" Trong h\u00ecnh d\u01b0\u1edbi, x'x v\u00e0 y'y l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng giao nhau t\u1ea1i O v\u00e0 $ \\widehat{xOy'} - m = 120^{o} $ <br\/> th\u00ec m + n b\u1eb1ng: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-17.png' \/><\/center> ","select":["A. $30^{0}$","B. $90^{0}$ ","C. $60^{0}$","D. $110^{0}$"],"hint":" <br\/> B1. T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{x'Oy'} = m$ <br\/> B2. T\u00ednh m + n = ? ","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh th\u00ec b\u1eb1ng nhau. <br\/> Hai g\u00f3c k\u1ec1 b\u00f9 c\u00f3 t\u1ed5ng b\u1eb1ng $180^{o}.$ <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> Ta c\u00f3: $\\widehat{xOy'} + \\widehat{x'Oy'} = 180^{o} $ (1) (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $ \\widehat{xOy'} - m = 120^{0} $ (2) (gi\u1ea3 thi\u1ebft)<br\/> T\u1eeb (1) v\u00e0 (2) suy ra $ \\begin{cases} \\widehat{xOy'} + m = 180 ^{o} \\\\ \\widehat{x'Oy} - m = 120^{0} \\end{cases} $ <br\/> $ \\Rightarrow m + m = 180^{o} - 120^{o} \\\\ \\Leftrightarrow 2m = 60^{o} \\\\ \\Leftrightarrow m = 30^{o} = \\widehat{x'Oy'}$ <br\/> M\u00e0 $ \\widehat{x'Oy'} = \\widehat{xOy} $ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) <br\/> $ \\Rightarrow m + n = 2m = 2.30^{o} = 60^{o} $<\/span> <br\/><br\/><span class='basic_pink'> \u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><\/span> ","column":4}]}],"id_ques":1359},{"time":14,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" Trong h\u00ecnh d\u01b0\u1edbi. C\u00f3 AC v\u00e0 MN l\u00e0 hai \u0111o\u1ea1n th\u1eb3ng. <br\/> G\u00f3c $ \\widehat{NBC} $ b\u1eb1ng: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai1/lv3/img\/H7C11-16.png' \/><\/center> ","select":["A. $36^{0}$","B. $72^{0}$ ","C. $38^{0}$","D. $75^{0}$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh th\u00ec b\u1eb1ng nhau. <br\/> Hai g\u00f3c k\u1ec1 b\u00f9 c\u00f3 t\u1ed5ng b\u1eb1ng $180^{o}$ <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <span class='basic_left'> Ta c\u00f3: $\\widehat{ABM} + \\widehat{ABN} = 180^{o} $ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow$ 2x + 3x = $180^{o}$<br\/>Hay 5x = $180^{o}$ $\\Rightarrow$ x = $36^{o}$<br\/> M\u00e0 $\\widehat{ABM}$ = 2x = 2. $36^{o}$ = $72^{o}$<br\/> L\u1ea1i c\u00f3: $\\widehat{ABM}$ = $\\widehat{NBC}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh)<br\/>$\\Rightarrow$ $\\widehat{NBC}$ = $72^{o}$<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>B. <\/span>$72^{o}$<\/span> <br\/> ","column":4}]}],"id_ques":1360}],"lesson":{"save":0,"level":3}}