{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/2.jpg' \/><\/center>$\\dfrac{7}{9}+\\dfrac{5}{12}- \\dfrac{3}{4}$ = <span class='basic_hoi_black'>?<\/span>","select":["A. $\\dfrac{3}{9}$","B. $\\dfrac{4}{9}$","C. $\\dfrac{5}{9}$","D. $\\dfrac{6}{9}$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n t\u00ednh nh\u01b0 sau:<br\/>+ B\u01b0\u1edbc 1: Quy \u0111\u1ed3ng m\u1eabu<br\/>+ B\u01b0\u1edbc 2: Th\u1ef1c hi\u1ec7n t\u00ednh t\u1eeb tr\u00e1i qua ph\u1ea3i<\/span> <br\/><br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\dfrac{7}{9}+\\dfrac{5}{12}- \\dfrac{3}{4}$<br\/>= $\\dfrac{28}{36}+\\dfrac{15}{36}- \\dfrac{27}{36}$<br\/>= $\\dfrac{43}{36}- \\dfrac{27}{36}$<br\/>= $\\dfrac{16}{36}$<br\/>= $\\dfrac{4}{9}$<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>B. $\\dfrac{4}{9}$<\/span><\/span> <br\/>","column":2}],"id_ques":2271},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" T\u00ecm x, bi\u1ebft: $\\dfrac{3}{8}$ - $\\dfrac{1}{6}$x = $\\dfrac{1}{4}$ ","select":["A. x = $\\dfrac{3}{2}$","B. x = $\\dfrac{1}{3}$","C. x = $\\dfrac{4}{3}$","D. x = $\\dfrac{3}{4}$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>+ B\u01b0\u1edbc 1: T\u00ecm $\\dfrac{1}{6}$x<br\/>+ B\u01b0\u1edbc 2: T\u00ecm x<\/span> <br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\dfrac{3}{8}$ - $\\dfrac{1}{6}$x = $\\dfrac{1}{4}$<br\/>$\\dfrac{1}{6}$x = $\\dfrac{3}{8}$ - $\\dfrac{1}{4}$<br\/>$\\dfrac{1}{6}$x = $\\dfrac{3}{8}$ - $\\dfrac{2}{8}$<br\/>$\\dfrac{1}{6}$x = $\\dfrac{1}{8}$<br\/>x = $\\dfrac{1}{8}$ : $\\dfrac{1}{6}$<br\/>x = $\\dfrac{1}{8}$ . 6<br\/>x = $\\dfrac{6}{8}$<br\/>x = $\\dfrac{3}{4}$<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D. x = $\\dfrac{3}{4}$<\/span> <br\/><\/span>","column":2}],"id_ques":2272},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" L\u1edbp 6A c\u00f3 45 h\u1ecdc sinh. Trong \u0111\u00f3 s\u1ed1 h\u1ecdc sinh trung b\u00ecnh chi\u1ebfm $\\dfrac{6}{15}$ s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp. T\u1ed5ng s\u1ed1 h\u1ecdc sinh kh\u00e1 v\u00e0 gi\u1ecfi chi\u1ebfm $1\\dfrac{1}{9}$ s\u1ed1 h\u1ecdc sinh trung b\u00ecnh, c\u00f2n l\u1ea1i l\u00e0 h\u1ecdc sinh y\u1ebfu k\u00e9m. T\u00ednh s\u1ed1 h\u1ecdc sinh y\u1ebfu k\u00e9m c\u1ee7a l\u1edbp 6A. ","select":["A. 6 h\u1ecdc sinh","B. 7 h\u1ecdc sinh","C. 8 h\u1ecdc sinh","D. 9 h\u1ecdc sinh"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm s\u1ed1 h\u1ecdc sinh trung b\u00ecnh<br\/>B\u01b0\u1edbc 2: T\u00ecm s\u1ed1 h\u1ecdc sinh kh\u00e1 v\u00e0 gi\u1ecfi<br\/>B\u01b0\u1edbc 3: T\u00ecm s\u1ed1 h\u1ecdc sinh y\u1ebfu k\u00e9m<\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$1\\dfrac{1}{9}$ = $\\dfrac{10}{9}$<br\/>S\u1ed1 h\u1ecdc sinh trung b\u00ecnh c\u1ee7a l\u1edbp 6A l\u00e0:<br\/>$\\dfrac{6}{15}$ . 45 = 18 (h\u1ecdc sinh)<br\/>S\u1ed1 h\u1ecdc sinh kh\u00e1 v\u00e0 gi\u1ecfi c\u1ee7a l\u1edbp 6A l\u00e0:<br\/>$\\dfrac{10}{9}$ . 18 = 20 ( h\u1ecdc sinh)<br\/>S\u1ed1 h\u1ecdc sinh y\u1ebfu k\u00e9m c\u1ee7a l\u1edbp 6A l\u00e0: 45 - (18 + 20) = 7 ( h\u1ecdc sinh)<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>B. 7 h\u1ecdc sinh<\/span><\/span> <br\/>","column":2}],"id_ques":2273},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/2.jpg' \/><\/center>T\u00ecm n, bi\u1ebft: (n + 4) $\\vdots$ (n + 1)","select":["A. n $\\in$ {4; -2; 0; 2}","B. n $\\in$ {-4; -2; 0; 2}","C. n $\\in$ { -4; -2; 1; 2}","D. n $\\in$ {-4; -2; 0; 4}"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch n + 4 v\u1ec1 d\u1ea1ng a.(n + 1) + b, (a, b $\\in $ Z, a $\\ne$ 0)<br\/>B\u01b0\u1edbc 2: T\u00ecm n<\/span> <br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: n + 4 = (n + 1) + 3<br\/>M\u00e0 (n + 4) $\\vdots$ (n + 1)<br\/>$\\Rightarrow$ (n + 1) + 3 $\\vdots$ (n + 1)<br\/>L\u1ea1i c\u00f3: (n + 1) $\\vdots$ (n + 1) v\u1edbi m\u1ecdi n<br\/> $\\Rightarrow$ 3 $\\vdots$ (n + 1) hay (n + 1) $\\in$ \u01af(3)<br\/>\u01af(3) = {-3; -1; 1; 3}<br\/> $\\Rightarrow$ (n + 1) $\\in$ {-3; -1; 1; 3}<br\/>$\\Rightarrow$ n $\\in$ {-4; -2; 0; 2}<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'> B. n $\\in$ {-4; -2; 0; 2}<\/span><\/span> <br\/><\/span><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t: <\/span><b> N\u1ebfu a + b chia h\u1ebft cho c v\u00e0 a chia h\u1ebft cho c th\u00ec b chia h\u1ebft cho c <\/b><\/span>","column":2}],"id_ques":2274},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/2.jpg' \/><\/center>$\\dfrac{-7}{9} . \\dfrac{5}{8} - \\dfrac{2}{9} : \\dfrac{8}{5}$ = <span class='basic_hoi_black'>?<\/span>","select":["A. $\\dfrac{8}{9}$","B. $\\dfrac{5}{8}$","C. $\\dfrac{-5}{8}$","D. $\\dfrac{9}{8}$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Bi\u1ec3u th\u1ee9c ch\u1ee9a ph\u00e9p t\u00ednh nh\u00e2n, chia v\u00e0 c\u1ed9ng, tr\u1eeb th\u00ec ta th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh nh\u00e2n, chia tr\u01b0\u1edbc, th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh c\u1ed9ng, tr\u1eeb sau.<\/span> <br\/><br\/><span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\dfrac{-7}{9} . \\dfrac{5}{8} - \\dfrac{2}{9} : \\dfrac{8}{5}$ <br\/>= $\\dfrac{-7}{9} . \\dfrac{5}{8} - \\dfrac{2}{9} . \\dfrac{5}{8}$<br\/>= $(\\dfrac{-7}{9} - \\dfrac{2}{9}) . \\dfrac{5}{8}$<br\/>= $\\dfrac{-9}{9} . \\dfrac{5}{8}$<br\/>= -1. $\\dfrac{5}{8}$<br\/>= $\\dfrac{-5}{8}$<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>C. $\\dfrac{-5}{8}$<\/span><\/span> <br\/>","column":2}],"id_ques":2275},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" T\u00ednh t\u1ed5ng:<br\/>C = $\\dfrac{4}{2.4} + \\dfrac{4}{4.6}+ \\dfrac{4}{6.8} + ... + \\dfrac{4}{2008.2010}$","select":["C = $\\dfrac{502}{1005}$","C = $\\dfrac{500}{1005}$","C = $\\dfrac{103}{1005}$","C = $\\dfrac{1004}{1005}$"],"hint":"$\\dfrac{4}{2.4}$ = 2.($\\dfrac{1}{2}$ - $\\dfrac{1}{4}$)","explain":" <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>C = $\\dfrac{4}{2.4} + \\dfrac{4}{4.6}+ \\dfrac{4}{6.8} + ... + \\dfrac{4}{2008.2010}$<br\/>C = 2. $(\\dfrac{2}{2.4} + \\dfrac{2}{4.6}+ \\dfrac{2}{6.8} + ... + \\dfrac{2}{2008.2010})$<br\/>C = 2. $(\\dfrac{1}{2}$ - $\\dfrac{1}{4}$ + $\\dfrac{1}{4}$ - $\\dfrac{1}{6}$ + $\\dfrac{1}{6}$ - $\\dfrac{1}{8}$ + ... - $\\dfrac{1}{2008}$ + $\\dfrac{1}{2008}$ - $\\dfrac{1}{2010})$<br\/>C = 2. $[\\dfrac{1}{2}$ + $(\\dfrac{-1}{4}$ + $\\dfrac{1}{4})$ + $(\\dfrac{-1}{6}$ + $\\dfrac{1}{6})$ + $(\\dfrac{-1}{8}$ + $\\dfrac{1}{8})$ + ... + $(\\dfrac{-1}{2008}$ + $\\dfrac{1}{2008})$ - $\\dfrac{1}{2010}]$<br\/>C = 2. $[\\dfrac{1}{2}$ + 0 + 0 + 0 + ... + 0 - $\\dfrac{1}{2010}]$<br\/>C = 2. $(\\dfrac{1}{2}$ - $\\dfrac{1}{2010})$<br\/>C = 2. $(\\dfrac{1005}{2010}$ - $\\dfrac{1}{2010})$<br\/>C = 2. $\\dfrac{1004}{2010}$<br\/>C = $\\dfrac{1004}{1005}$<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>C = $\\dfrac{1004}{1005}$<\/span><\/span> <br\/><\/span><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t: <\/span>$\\dfrac{b - a}{a.b}$ = $\\dfrac{1}{a}$ - $\\dfrac{1}{b}$<\/span>","column":2}],"id_ques":2276},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/H628_D101.png' \/><\/center> Cho h\u00ecnh v\u1ebd c\u00f3 s\u1ed1 \u0111o nh\u01b0 tr\u00ean. T\u00ednh $\\widehat{yOz}$. ","select":["A. $\\widehat{yOz}$ = $20^o$","B. $\\widehat{yOz}$ = $30^o$","C. $\\widehat{yOz}$ = $45^o$","D. $\\widehat{yOz}$ = $60^o$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/H628_D101.png' \/><\/center>Ta c\u00f3: $\\widehat{xOy}$ < $\\widehat{xOz}$ ( v\u00ec $75^o$ < $120^o$)<br\/> N\u00ean tia Oy n\u1eb1m gi\u1eefa hai tia Ox v\u00e0 Oz<br\/>Suy ra $\\widehat{xOy}$ + $\\widehat{yOz}$ = $\\widehat{xOz}$<br\/>Thay $\\widehat{xOy}$ = $75^o$, $\\widehat{xOz}$ = $120^o$, ta \u0111\u01b0\u1ee3c:<br\/> $75^o$ + $\\widehat{yOz}$ = $120^o$ <br\/>Suy ra $\\widehat{yOz}$ = $120^o$ - $75^o$ = $45^o$ <br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>C. <\/span>$\\widehat{yOz}$ = $45^o$<\/span> <br\/><\/span><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> $\\widehat{xOy}$ = $m^o$, $\\widehat{xOz}$ = $n^o$. N\u1ebfu $m^o$ < $n^o$ th\u00ec tia Oy n\u1eb1m gi\u1eefa hai tia Ox v\u00e0 Oz<\/span>","column":2}],"id_ques":2277},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["54"],["108"]]],"list":[{"point":10,"width":60,"type_input":"","ques":"Cho tia Oy l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c xOz. Bi\u1ebft $\\widehat{xOy}$ = $54^o$. V\u1eady $\\widehat{yOz}$ = _input_$^o$, $\\widehat{xOz}$ = _input_$^o$","hint":"","explain":"<span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/H628_D104.png' \/><\/center>Ta c\u00f3: Tia Oy l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{xOz}$ <br\/>Suy ra $\\widehat{xOy}$ = $\\widehat{yOz}$ = $\\frac{1}{2}$ $\\widehat{xOz}$ = $54^o$<br\/>L\u1ea1i c\u00f3: $\\widehat{xOy}$ = $\\frac{1}{2}$ $\\widehat{xOz}$<br\/>Do \u0111\u00f3: $\\widehat{xOz}$ = 2. $\\widehat{xOy}$ = 2. $54^o$ = $108^o$<br\/>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t t\u1eeb tr\u00e1i qua ph\u1ea3i l\u00e0: <span class='basic_pink'> 54, 108<\/span><\/span> <br\/> "}],"id_ques":2278},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" Cho hai tia Ox, Oy c\u00f9ng n\u1eb1m tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng c\u00f3 b\u1edd ch\u1ee9a tia Ot. Bi\u1ebft $\\widehat{tOy}$ = $80^o$, $\\widehat{xOy}$ = $30^o$ v\u00e0 tia Oy n\u1eb1m gi\u1eefa hai tia Ox v\u00e0 Ot. G\u1ecdi On l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c xOy, Om l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c tOy. S\u1ed1 \u0111o c\u1ee7a g\u00f3c mOn l\u00e0:","select":["A. $40^o$","B. $70^o$","C. $60^o$","D. $55^o$"],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/H628_D105.png' \/><\/center> Ta c\u00f3: Tia Oy n\u1eb1m gi\u1eefa hai tia Ox v\u00e0 Ot<br\/>Do \u0111\u00f3: $\\widehat{tOy}$ + $\\widehat{xOy}$ = $\\widehat{xOt}$ <br\/>Thay $\\widehat{tOy}$ = $80^o$, $\\widehat{xOy}$ = $30^o$ ta \u0111\u01b0\u1ee3c: <br\/>$\\widehat{xOt}$ = $80^o$ + $30^o$ = $110^o$<br\/>+ Om l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c tOy<br\/>Do \u0111\u00f3: $\\widehat{tOm}$ = $\\widehat{mOy}$ = $\\dfrac{1}{2}$ $\\widehat{tOy}$ = $\\dfrac{1}{2}$.$80^o$ = $40^o$<br\/>+ Tia On l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c xOy<br\/>Do \u0111\u00f3: $\\widehat{yOn}$ = $\\widehat{xOn}$ = $\\dfrac{1}{2}$ $\\widehat{xOy}$ = $\\dfrac{1}{2}$.$30^o$ = $15^o$<br\/>+ $\\widehat{xOn}$ < $\\widehat{xOt}$ ( v\u00ec $15^o$ < $110^o$) n\u00ean tia On n\u1eb1m gi\u1eefa hai tia Ox v\u00e0 Ot<br\/>Suy ra: $\\widehat{xOn}$ + $\\widehat{nOt}$ = $\\widehat{xOt}$<br\/>Thay $\\widehat{xOn}$ = $15^o$, $\\widehat{xOt}$ = $110^o$ ta \u0111\u01b0\u1ee3c: <br\/>$15^o$ + $\\widehat{nOt}$ = $110^o$<br\/>$\\widehat{nOt}$ = $110^o$ - $15^o$ = $95^o$<br\/>+ $\\widehat{tOm}$ < $\\widehat{tOn}$ ( v\u00ec $40^o$ < $95^o$) n\u00ean tia Om n\u1eb1m gi\u1eefa hai tia Ot v\u00e0 On<br\/>Suy ra: $\\widehat{tOm}$ + $\\widehat{mOn}$ = $\\widehat{tOn}$<br\/>Thay $\\widehat{tOm}$ = $40^o$, $\\widehat{tOn}$ = $95^o$ ta \u0111\u01b0\u1ee3c:<br\/>$40^o$ + $\\widehat{mOn}$ = $95^o$<br\/>$\\widehat{mOn}$ = $95^o$ - $40^o$ = $55^o$<br\/>V\u1eady ta ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>D. $55^o$<\/span><\/span> <br\/><\/span><\/span><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Oy l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c xOz khi:<br\/>+ Tia Oy l\u00e0 tia n\u1eb1m gi\u1eefa hai tia Ox v\u00e0 Oz<br\/>+ $\\widehat{xOy}$ = $\\widehat{yOz}$ = $\\dfrac{1}{2}$$\\widehat{xOz}$<\/span>","column":2}],"id_ques":2279},{"title":"\u0110i\u1ec1n s\u1ed1 ho\u1eb7c ch\u1eef th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["C"],["4"]]],"list":[{"point":10,"width":60,"type_input":"","ques":"<span class='basic_left'>V\u1ebd tam gi\u00e1c ABC, bi\u1ebft ba c\u1ea1nh BC = 8cm, AB = 4cm, AC = 6cm.<br\/><b>C\u00e1ch v\u1ebd:<\/b><br\/>+ V\u1ebd \u0111o\u1ea1n th\u1eb3ng BC = 8cm<br\/>+ V\u1ebd cung tr\u00f2n t\u00e2m _input_, b\u00e1n k\u00ednh 6cm<br\/>+ V\u1ebd cung tr\u00f2n t\u00e2m B, b\u00e1n k\u00ednh _input_cm.<br\/>+ L\u1ea5y m\u1ed9t giao \u0111i\u1ec3m c\u1ee7a hai cung tr\u00ean, g\u1ecdi giao \u0111i\u1ec3m \u0111\u00f3 l\u00e0 A<br\/>+ V\u1ebd \u0111o\u1ea1n th\u1eb3ng AB, AC, ta c\u00f3 $\\triangle$ABC.<\/span>","hint":"","explain":"<span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop6/toan/daiso/kiemtracuoinam/lv3/img\/H628_D106.png' \/><\/center><b>C\u00e1ch v\u1ebd:<\/b><br\/>+ V\u1ebd \u0111o\u1ea1n th\u1eb3ng BC = 8cm<br\/>+ V\u1ebd cung tr\u00f2n t\u00e2m C, b\u00e1n k\u00ednh 6cm<br\/>+ V\u1ebd cung tr\u00f2n t\u00e2m B, b\u00e1n k\u00ednh 4cm.<br\/>+ L\u1ea5y m\u1ed9t giao \u0111i\u1ec3m c\u1ee7a hai cung tr\u00ean, g\u1ecdi giao \u0111i\u1ec3m \u0111\u00f3 l\u00e0 A<br\/>+ V\u1ebd \u0111o\u1ea1n th\u1eb3ng AB, AC, ta c\u00f3 $\\triangle$ABC.<br\/>V\u1eady s\u1ed1 ho\u1eb7c ch\u1eef c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng theo th\u1ee9 t\u1ef1 t\u1eeb tr\u00ean xu\u1ed1ng d\u01b0\u1edbi l\u00e0:<span class='basic_pink'> C; 4<\/span><\/span> <\/span> <br\/> "}],"id_ques":2280}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}