{"common":{"save":0,"post_id":"818","level":2,"total":20,"point":5,"point_extra":0},"segment":[{"id":"1771","post_id":"818","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["185"],["4"]]],"list":[{"point":5,"width":50,"ques":"T\u00ecm x bi\u1ebft $\\left( \\dfrac{2}{3}x - \\dfrac{1}{2} \\right) . \\dfrac{3}{4} - \\dfrac{2}{5}x = 4\\dfrac{1}{4}$ <br\/> V\u1eady $x$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div> ","hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Nh\u00e2n ph\u00e1 ngo\u1eb7c <br\/> <b> B\u01b0\u1edbc 2: <\/b> chuy\u1ec3n c\u00e1c s\u1ed1 h\u1ea1ng kh\u00f4ng ch\u1eefa x sang v\u1ebf ph\u1ea3i <br\/> <b> B\u01b0\u1edbc 3: <\/b> Bi\u1ebfn \u0111\u1ed5i v\u00e0 t\u00ecm x <\/span> <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $ \\begin{align*} \\left( \\dfrac{2}{3}x - \\dfrac{1}{2} \\right) . \\dfrac{3}{4} - \\dfrac{2}{5}x &= 4\\dfrac{1}{4} \\\\ \\dfrac{1}{2}x - \\dfrac{3}{8} - \\dfrac{2}{5}x &= \\dfrac{17}{4} \\\\ \\dfrac{1}{2}x - \\dfrac{2}{5}x &= \\dfrac{17}{4} + \\dfrac{3}{8} \\\\ \\dfrac{1}{10}x &= \\dfrac{37}{8} \\\\ x &= \\dfrac{37}{8} : \\dfrac{1}{10} \\\\ x &= \\dfrac{185}{4}\\end{align*} $"}]}]},"correct":"","level":"2","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:55"},{"id":"1772","post_id":"818","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"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":"T\u00ecm x $\\in \\mathbb{Z}$, bi\u1ebft $\\dfrac{-2}{3} + \\dfrac{-5}{12} < x \\leq 4 - \\dfrac{1}{3} : \\dfrac{1}{6}$ ","select":["A. $x \\in \\left \\{ -1; 0; 1; 2 \\right\\}$ ","B. $x \\in \\left\\{ -2; -1; 0; 1 \\right\\}$ ","C. $x \\in \\left\\{ -4; -3; -2; -1; 0 \\right\\}$ ","D. $x \\in \\left\\{ 0; 1; 2; 3; 4 \\right\\}$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ednh bi\u1ec3u th\u1ee9c v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i sau \u0111\u00f3 t\u00ecm x <\/span> <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> Ta c\u00f3: <br\/> $\\dfrac{-2}{3} + \\dfrac{-5}{12} = \\dfrac{-8}{12} + \\dfrac{-5}{12} = \\dfrac{-13}{12}$ <br\/> $4 - \\dfrac{1}{3} : \\dfrac{1}{6} = 4 - \\dfrac{1}{3} . 6 = 4 - 2 = 2$ <br\/> $\\Rightarrow$ $\\dfrac{-13}{12} < x \\leq 2$ hay $-1\\dfrac{1}{12} < x \\leq 2$ <br\/> V\u00ec $x \\in \\mathbb{Z}$ $\\Rightarrow$ $x \\in \\left\\{ -1; 0; 1; 2 \\right\\}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":2}]}]},"correct":"","level":"2","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:55"},{"id":"1773","post_id":"818","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"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 p v\u00e0 2p + 1 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 (p > 5). <br\/> Lan kh\u1eb3ng \u0111\u1ecbnh 4p + 1 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 <b> \u0111\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"S\u1ed1 nguy\u00ean t\u1ed1 l\u00e0 s\u1ed1 ch\u1ec9 chia h\u1ebft cho 1 v\u00e0 ch\u00ednh n\u00f3","explain":"<span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> V\u00ec p l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1, p > 5 n\u00ean: p $\\not \\vdots$ 3 $\\Rightarrow$ 4p $\\not \\vdots$ 3 (1) <br\/> Do 2p + 1 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 l\u1edbn h\u01a1n 3 n\u00ean: 2p + 1 $\\not \\vdots$ 3 <br\/> $\\Rightarrow$ 2(2p + 1) $\\not \\vdots$ 3 hay 4p + 2 $\\not \\vdots$ 3 (1) <br\/> M\u1eb7t kh\u00e1c trong 3 s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp 4p, 4p + 1; 4p + 2 lu\u00f4n c\u00f3 1 s\u1ed1 chia h\u1ebft cho 3 (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ 4p + 1 $\\vdots$ 3 <br\/> M\u00e0 4p + 1 > 3 n\u00ean 4p + 1 l\u00e0 h\u1ee3p s\u1ed1 <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean c\u1ee7a Lan l\u00e0 <span class='basic_pink'> SAI <\/span> ","column":2}]}]},"correct":"","level":"2","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:55"}]}