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Home Lớp 6 Toán lớp 6 - Sách kết nối tri thức Phép trừ phân sốBài tập nâng cao

{"common":{"save":0,"post_id":"806","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1641","post_id":"806","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":[[["999"],["1000"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ednh nhanh <br\/> <br\/> $A = \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + \\dfrac{1}{3 . 4} + ... + \\dfrac{1}{999 . 1000} = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","hint":"\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c $\\dfrac{1}{n.(n - 1)} = \\dfrac{1}{n} - \\dfrac{1}{n + 1}$ $(n \\in \\mathbb{N^*})$ ","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c $\\dfrac{1}{n.(n - 1)} = \\dfrac{1}{n} - \\dfrac{1}{n + 1}$ \u0111\u1ec3 ph\u00e2n t\u00edch c\u00e1c ph\u00e2n s\u1ed1 tr\u00ean <br\/> B\u01b0\u1edbc 2: <\/b> Tri\u1ec7t ti\u00eau c\u00e1c ph\u00e2n s\u1ed1 \u0111\u1ed1i nhau <br\/> B\u01b0\u1edbc 3: <\/b> T\u00ecm k\u1ebft qu\u1ea3 cu\u1ed1i c\u00f9ng <\/span> <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} A &= \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + \\dfrac{1}{3 . 4} + ... + \\dfrac{1}{999 . 1000} \\\\ &= \\dfrac{2 - 1}{1 . 2} + \\dfrac{3 - 2}{2 . 3} + \\dfrac{4 - 3}{3 . 4} + ... + \\dfrac{1000 - 999}{999 . 1000} \\\\ &= \\dfrac{2}{1 . 2} - \\dfrac{1}{1 . 2} + \\dfrac{3}{2 . 3} - \\dfrac{2}{2 . 3} + \\dfrac{4}{3 . 4} - \\dfrac{3}{3 . 4} + ... + \\dfrac{1000}{999 . 1000} - \\dfrac{999}{999 . 1000} \\\\ &= 1 - \\dfrac{1}{2}+ \\dfrac{1}{2} - \\dfrac{1}{3} + \\dfrac{1}{3} - \\dfrac{1}{4} + ... + \\dfrac{1}{999} - \\dfrac{1}{1000} \\\\ &= 1 - \\dfrac{1}{1000} \\\\ &= \\dfrac{999}{1000} \\end{align*}$ <br\/> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 999; 1000 <\/span> <br\/> Nh\u1eadn x\u00e9t: C\u00f4ng th\u1ee9c $\\dfrac{1}{n.(n - 1)} = \\dfrac{1}{n} - \\dfrac{1}{n + 1}$ $(n \\in \\mathbb{N^*})$ gi\u00fap ta t\u00ednh nhanh \u0111\u01b0\u1ee3c t\u1ed5ng c\u00e1c ph\u00e2n s\u1ed1 vi\u1ebft theo quy lu\u1eadt v\u00ec \u0111\u00e3 l\u00e0m xu\u1ea5t hi\u1ec7n c\u00e1c s\u1ed1 \u0111\u1ed1i nhau <\/i> <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:53"},{"id":"1642","post_id":"806","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":"Cho $S = \\dfrac{1}{2^2} + \\dfrac{1}{3^2} + \\dfrac{1}{4^2} + \\dfrac{1}{5^2} + \\dfrac{1}{6^2}$ <br\/> <br\/>Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. S > 1 ","B. S = 1","C. S < 1"],"hint":"So s\u00e1nh $\\dfrac{1}{n^2}$ v\u1edbi $\\dfrac{1}{(n - 1) . n}$ ","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh m\u1ed7i ph\u00e2n s\u1ed1 c\u00f3 d\u1ea1ng $\\dfrac{1}{n^2}$ v\u1edbi ph\u00e2n s\u1ed1 $\\dfrac{1}{(n - 1) . n}$ <br\/> B\u01b0\u1edbc 2: <\/b> T\u1eeb b\u01b0\u1edbc 1 suy ra $S < \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + ... + \\dfrac{1}{5 . 6}$ <br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh t\u1ed5ng $\\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + ... + \\dfrac{1}{5 . 6}$ v\u00e0 so s\u00e1nh v\u1edbi 1 <\/span> <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <br\/> Ta c\u00f3: <br\/> $\\dfrac{1}{2^2} < \\dfrac{1}{1 . 2}$ <br\/> $\\dfrac{1}{3^2} < \\dfrac{1}{2 . 3}$ <br\/> ......... <br\/> $\\dfrac{1}{6^2} < \\dfrac{1}{5 . 6}$ <br\/> $\\Rightarrow$ $ S < \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + ... + \\dfrac{1}{5 . 6}$ <br\/> $\\begin{align*} & \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + ... + \\dfrac{1}{5 . 6} \\\\ &= 1 - \\dfrac{1}{2} + \\dfrac{1}{2} - \\dfrac{1}{3} + ... + \\dfrac{1}{5} - \\dfrac{1}{6} \\\\ &= 1 - \\dfrac{1}{6} \\\\ &= \\dfrac{5}{6} \\end{align*}$ <br\/> $\\Rightarrow$ $ S < \\dfrac{5}{6}$ v\u00e0 $\\dfrac{5}{6} < 1$ <br\/> N\u00ean: $S = \\dfrac{1}{2^2} + \\dfrac{1}{3^2} + \\dfrac{1}{4^2} + \\dfrac{1}{5^2} + \\dfrac{1}{6^2} < 1$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C. S < 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"},{"id":"1643","post_id":"806","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":10,"ques":"Cho $S = \\dfrac{1}{2} + \\dfrac{1}{6} + \\dfrac{1}{12} + \\dfrac{1}{20} + \\dfrac{1}{30} + \\dfrac{1}{42} + \\dfrac{1}{56} + \\dfrac{1}{72} + \\dfrac{1}{90}$ <br\/><br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. S < 1 ","B. S = 1","C. S > 1"],"hint":"Ph\u00e2n t\u00edch c\u00e1c m\u1eabu s\u1ed1 th\u00e0nh t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Nh\u1eadn th\u1ea5y c\u00e1c s\u1ed1 2 = 1 . 2; 6 = 2 . 3; 12 = 3 . 4.... <br\/> N\u00ean $S = \\dfrac{1}{2} + \\dfrac{1}{6} + \\dfrac{1}{12} + \\dfrac{1}{20} + \\dfrac{1}{30} + \\dfrac{1}{42} + \\dfrac{1}{56} + \\dfrac{1}{72} + \\dfrac{1}{90} \\\\ = \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + \\dfrac{1}{3 . 4} + \\dfrac{1}{4 . 5} + \\dfrac{1}{5 . 6} + ... + \\dfrac{1}{9 . 10}$ <br\/> T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c t\u1ed5ng S sau \u0111\u00f3 so s\u00e1nh v\u1edbi 1 <\/span> <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <br\/> Ta c\u00f3: <br\/> $\\begin{align*} S &= \\dfrac{1}{2} + \\dfrac{1}{6} + \\dfrac{1}{12} + \\dfrac{1}{20} + \\dfrac{1}{30} + \\dfrac{1}{42} + \\dfrac{1}{56} + \\dfrac{1}{72} + \\dfrac{1}{90} \\\\ &= \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + \\dfrac{1}{3 . 4} + \\dfrac{1}{4 . 5} + \\dfrac{1}{5 . 6} + \\dfrac{1}{6 . 7} + \\dfrac{1}{7 . 8} + \\dfrac{1}{8 . 9} + \\dfrac{1}{9 . 10} \\\\ &= 1 - \\dfrac{1}{2} + \\dfrac{1}{2} - \\dfrac{1}{3} + \\dfrac{1}{3} - \\dfrac{1}{4} + \\dfrac{1}{4} - \\dfrac{1}{5} + \\dfrac{1}{5} - \\dfrac{1}{6} + \\dfrac{1}{6} - \\dfrac{1}{7} + \\dfrac{1}{7} - \\dfrac{1}{8} + \\dfrac{1}{8} - \\dfrac{1}{9} + \\dfrac{1}{9} - \\dfrac{1}{10} \\\\ &= 1 - \\dfrac{1}{10} \\\\ &= \\dfrac{9}{10} \\end{align*}$ <br\/> V\u00ec $\\dfrac{9}{10} < 1$ n\u00ean S < 1 <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A. S < 1 <\/span> <br\/> Nh\u1eadn x\u00e9t: Kh\u00e9o l\u00e9o ph\u00e2n t\u00edch m\u1eabu s\u1ed1 th\u00e0nh t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb \u0111\u00f3 ph\u00e2n t\u00edch l\u00e0m xu\u1ea5t hi\u1ec7n c\u00e1c ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{1}{a} - \\dfrac{1}{b} + \\dfrac{1}{b} - \\dfrac{1}{c}$, tri\u1ec7t ti\u00eau c\u00e1c ph\u00e2n s\u1ed1 \u0111\u1ed1i, t\u1eeb \u0111\u00f3 ta t\u00ednh t\u1ed5ng m\u1ed9t c\u00e1ch d\u1ec5 d\u00e0ng <\/i> <\/span> ","column":3}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:54"}]}