Phép cộng phân số. Tính chất cơ bản của phép cộng phân số-level3

Home Lớp 6 Toán lớp 6 - Sách kết nối tri thức Phép cộng phân số. Tính chất cơ bản của phép cộng phân sốBài tập nâng cao

{"common":{"save":0,"post_id":"803","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1591","post_id":"803","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":[[3]],"list":[{"point":10,"ques":"T\u00ecm s\u1ed1 nguy\u00ean x bi\u1ebft: <br\/> <br\/> $ \\dfrac{1}{3} + \\dfrac{-2}{5} + \\dfrac{1}{6} + \\dfrac{-1}{5} \\leq x < \\dfrac{-3}{4} + \\dfrac{2}{7} + \\dfrac{-1}{4} + \\dfrac{3}{5} + \\dfrac{5}{7} $ <br\/> \u0110\u00e1p \u00e1n l\u00e0:","select":["A. $x \\in \\left\\{ 1; 2; 3; 4; 5; 6; 7 \\right\\}$ ","B. $x \\in \\left\\{ 7; 6; 5; 4; 3 \\right\\}$ ","C. $x \\in \\left\\{ -1; 0; 1; 2; 3; 4; 5 \\right\\}$","D. $x \\in \\left\\{-6; -5; -4; -3; -2; -1 \\right\\}$ "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh $ \\dfrac{1}{3} + \\dfrac{-2}{5} + \\dfrac{1}{6} + \\dfrac{-1}{5}$ b\u1eb1ng c\u00e1ch nh\u00f3m c\u00e1c ph\u00e2n s\u1ed1 m\u1ed9t c\u00e1ch th\u00edch h\u1ee3p <br\/> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\dfrac{-3}{4} + \\dfrac{2}{7} + \\dfrac{-1}{4} + \\dfrac{3}{5} + \\dfrac{5}{7} $ b\u1eb1ng c\u00e1ch nh\u00f3m c\u00e1c ph\u00e2n s\u1ed1 c\u00f9ng m\u1eabu <br\/> B\u01b0\u1edbc 3: <\/b> Quy \u0111\u1ed3ng m\u1eabu s\u1ed1 c\u00e1c ph\u00e2n s\u1ed1 v\u00e0 t\u00ecm s\u1ed1 nguy\u00ean x th\u1ecfa m\u00e3n y\u00eau c\u1ea7u c\u1ee7a b\u00e0i to\u00e1n <\/span> <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> $\\begin{align*} & \\dfrac{1}{3} + \\dfrac{-2}{5} + \\dfrac{1}{6} + \\dfrac{-1}{5} \\\\ &= \\left( \\dfrac{-2}{5} + \\dfrac{-1}{5} \\right) + \\dfrac{1}{3} + \\dfrac{1}{6} \\\\ &= \\dfrac{-2 + (-1)}{5} + \\dfrac{2}{6} + \\dfrac{1}{6} \\\\ &= \\dfrac{-3}{5} + \\dfrac{2 + 1}{6} \\\\ &= \\dfrac{-3}{5} + \\dfrac{1}{2} \\\\ &= \\dfrac{-3 . 2}{5 . 2} + \\dfrac{1 . 5}{2 . 5} \\\\ &= \\dfrac{-6 + 5}{10} \\\\ &= \\dfrac{-1}{10} \\end{align*}$ <br\/> $\\begin{align*} & \\dfrac{-3}{4} + \\dfrac{2}{7} + \\dfrac{-1}{4} + \\dfrac{3}{5} + \\dfrac{5}{7} \\\\ &=\\left( \\dfrac{-3}{4} + \\dfrac{-1}{4} \\right) + \\left( \\dfrac{2}{7} + \\dfrac{5}{7} \\right) + \\dfrac{3}{5} \\\\ &= \\dfrac{-1 + (-3)}{4} + \\dfrac{2 + 5}{7} + \\dfrac{3}{5} \\\\ &= -1 + 1 + \\dfrac{3}{5} \\\\ &= \\dfrac{3}{5} = \\dfrac{6}{10} \\end{align*}$ <br\/> $\\Rightarrow$ $\\dfrac{-1}{10} \\leq x < \\dfrac{6}{10}$ <br\/> $\\Rightarrow$ $x \\in$ {-1; 0; 1; 2; 3; 4; 5} <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> ","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:53"},{"id":"1592","post_id":"803","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":[[4]],"list":[{"point":10,"ques":"T\u00ecm s\u1ed1 nguy\u00ean x bi\u1ebft: <br\/> <br\/> $ -1 + \\dfrac{1}{3} + \\dfrac{2}{9} < x < \\dfrac{-1}{3} + \\dfrac{-4}{9} + \\dfrac{8}{9}$ <br\/> \u0110\u00e1p \u00e1n l\u00e0:","select":["A. $x \\in \\left\\{-4; -3; -2; -1; 0; 1 \\right\\}$ ","B. $x \\in \\left\\{ -3, -2, -1, 0, 1 \\right\\}$ ","C. $x \\in \\left\\{ -4, -3, -2, -1, 0 \\right\\}$","D. $x \\in \\left\\{-3, -2, -1, 0 \\right\\}$ "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh $ -1 + \\dfrac{1}{3} + \\dfrac{2}{9}$ <br\/> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\dfrac{-1}{3} + \\dfrac{-4}{9} + \\dfrac{8}{9}$ <br\/> B\u01b0\u1edbc 3: <\/b> Quy \u0111\u1ed3ng m\u1eabu s\u1ed1 c\u00e1c ph\u00e2n s\u1ed1 v\u00e0 t\u00ecm s\u1ed1 nguy\u00ean x th\u1ecfa m\u00e3n y\u00eau c\u1ea7u c\u1ee7a b\u00e0i to\u00e1n <\/span> <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> $\\begin{align*} & -1 + \\dfrac{1}{3} + \\dfrac{2}{9} \\\\ &= \\dfrac{-9}{9} + \\dfrac{1 . 3}{3 . 3} + \\dfrac{2}{9} \\\\ &= \\dfrac{-9}{9} + \\dfrac{3}{9} + \\dfrac{2}{9} \\\\ &= \\dfrac{-9 + 3 + 2}{9} \\\\ &= \\dfrac{-4}{9} \\end{align*}$ <br\/> $\\begin{align*} & \\dfrac{-1}{3} + \\dfrac{-4}{9} + \\dfrac{8}{9} \\\\ &= \\dfrac{-1 . 3}{3 . 3} + \\dfrac{-4}{9} + \\dfrac{8}{9} \\\\ &= \\dfrac{-3}{9} + \\dfrac{-4}{9} + \\dfrac{8}{9} \\\\ &= \\dfrac{-3 + (-4) + 8}{9} \\\\ &= \\dfrac{1}{9} \\end{align*}$ <br\/> $\\Rightarrow$ $\\dfrac{-4}{9} < x < \\dfrac{1}{9}$ <br\/> $\\Rightarrow$ $x \\in$ {-3, -2, -1, 0} <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:53"},{"id":"1593","post_id":"803","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":10,"ques":"Cho $A = \\dfrac{1}{10} + \\dfrac{1}{11} + \\dfrac{1}{12} + ... + \\dfrac{1}{18} + \\dfrac{1}{19}$ <br\/> <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang?","select":["A. A > 1 ","B. A < 1 ","C. A = 1"],"hint":"So s\u00e1nh t\u1eebng ph\u00e2n s\u1ed1 trong t\u1ed5ng v\u1edbi $\\dfrac{1}{10}$ ","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh t\u1eebng s\u1ed1 h\u1ea1ng trong t\u1ed5ng v\u1edbi $\\dfrac{1}{10}$ <br\/> B\u01b0\u1edbc 2: <\/b> V\u00ec t\u1ed5ng tr\u00ean c\u00f3 10 ph\u00e2n s\u1ed1 n\u00ean so s\u00e1nh <br\/> $\\dfrac{1}{10} + \\dfrac{1}{11} + \\dfrac{1}{12} + ... + \\dfrac{1}{18} + \\dfrac{1}{19}$ <br\/>V\u1edbi $\\underbrace{ \\dfrac{1}{10} + \\dfrac{1}{10} + \\dfrac{1}{10} + ... + \\dfrac{1}{10} + \\dfrac{1}{10}}_{10 \\text{s\u1ed1 h\u1ea1ng}} $ <br\/> B\u01b0\u1edbc 3: <\/b> Bi\u1ebfn \u0111\u1ed5i v\u00e0 k\u1ebft lu\u1eadn <\/span> <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> Ta c\u00f3: <br\/> $\\dfrac{1}{11} < \\dfrac{1}{10}$ <br\/> $\\dfrac{1}{12} < \\dfrac{1}{10}$ <br\/> ............... <br\/> $\\dfrac{1}{18} < \\dfrac{1}{10}$ <br\/> $\\dfrac{1}{19} < \\dfrac{1}{10}$ <br\/> V\u00ec t\u1ed5ng tr\u00ean c\u00f3 10 ph\u00e2n s\u1ed1 n\u00ean: <br\/> $\\dfrac{1}{10} + \\dfrac{1}{11} + \\dfrac{1}{12} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} < \\underbrace{ \\dfrac{1}{10} + \\dfrac{1}{10} + \\dfrac{1}{10} + ... + \\dfrac{1}{10} + \\dfrac{1}{10}}_{10 \\text{s\u1ed1 h\u1ea1ng}} $ <br\/> M\u00e0 $\\underbrace{ \\dfrac{1}{10} + \\dfrac{1}{10} + \\dfrac{1}{10} + ... + \\dfrac{1}{10} + \\dfrac{1}{10}}_{10 \\text{s\u1ed1 h\u1ea1ng}} = \\dfrac{1 + 1 + ... + 1}{10} = \\dfrac{10}{10} = 1$ <br\/> $\\Rightarrow$ $\\dfrac{1}{10} + \\dfrac{1}{11} + \\dfrac{1}{12} + ... + \\dfrac{1}{18} + \\dfrac{1}{19}$ < 1 hay A < 1 <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A < 1 <\/span> ","column":3}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:53"}]}