Phép nhân và phép chia phận số. Tính chất cơ bản của phép nhân phân số-level3

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

{"common":{"save":0,"post_id":"809","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1691","post_id":"809","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":[[["1"],["20"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"R\u00fat g\u1ecdn. <br\/> <br\/> $A = \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{19}{1} + \\dfrac{18}{2} + \\dfrac{17}{3} + ... + \\dfrac{2}{18} + \\dfrac{1}{19} } = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","hint":"","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft t\u1eed s\u1ed1 c\u1ee7a c\u00e1c ph\u00e2n s\u1ed1 $\\dfrac{19}{1}; \\dfrac{18}{2}; .... ; \\dfrac{2}{18}; \\dfrac{1}{19}$ th\u00e0nh hi\u1ec7u c\u1ee7a 20 v\u00e0 m\u1ed9t s\u1ed1 (V\u00ed d\u1ee5 $\\dfrac{19}{1} = \\dfrac{20 - 1}{1}$) <br\/> B\u01b0\u1edbc 2: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 v\u1eeba vi\u1ebft \u1edf b\u01b0\u1edbc 1 th\u00e0nh hi\u1ec7u 2 ph\u00e2n s\u1ed1 (V\u00ed d\u1ee5: $\\dfrac{20 - 1}{1} = \\dfrac{20}{1} - 1$) <br\/> B\u01b0\u1edbc 3: <\/b> \u0110\u1eb7t nh\u00e2n t\u1eed chung ra ngo\u00e0i v\u00e0 r\u00fat g\u1ecdn <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} A &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{19}{1} + \\dfrac{18}{2} + \\dfrac{17}{3} + ... + \\dfrac{2}{18} + \\dfrac{1}{19} } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{20 - 1}{1} + \\dfrac{20 - 2}{2} + \\dfrac{20 - 3}{3} + ... + \\dfrac{20 - 18}{18} + \\dfrac{20 - 19}{19} } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\dfrac{20}{1} - 1 + \\dfrac{20}{2} - 1 + \\dfrac{20}{3} - 1 + ... + \\dfrac{20}{18} - 1 + \\dfrac{20}{19} - 1} \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ \\left( \\dfrac{20}{1} + \\dfrac{20}{2} + \\dfrac{20}{3} + ... + \\dfrac{20}{18} + \\dfrac{20}{19} \\right) - \\left( \\underbrace{1 + 1 + ... + 1}_{19 \\hspace{0.3cm} \\text{s\u1ed1 1}} \\right) } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ 20 + 20 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} \\right) - 19 } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ 20 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} \\right) + 20 . \\dfrac{1}{20} } \\\\ &= \\dfrac{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{19} + \\dfrac{1}{20} }{ 20 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{18} + \\dfrac{1}{19} + \\dfrac{1}{20} \\right)} \\\\ &= \\dfrac{1}{20} \\end{align*}$ <br\/> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 1; 20 <\/span> <br\/> Nh\u1eadn x\u00e9t: V\u1edbi nh\u1eefng b\u00e0i r\u00fat g\u1ecdn nh\u01b0 tr\u00ean ta ch\u1ec9 c\u1ea7n x\u00e9t m\u1eabu s\u1ed1 <br\/> $\\dfrac{n + m - 1}{1} + \\dfrac{n + m - 2}{2} + \\dfrac{n + m - 3}{3} + ... + \\dfrac{2}{n + m - 2} + \\dfrac{1}{n + m - 1} \\\\ = \\dfrac{n + m}{1} - 1 + \\dfrac{n + m}{2} - 1 + \\dfrac{n + m}{3} - 1 + ... + \\dfrac{n + m}{n + m - 2} - 1 + \\dfrac{n + m }{n + m - 1} - 1 \\\\ = (n + m) . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{n + m - 1} \\right) + 1 \\\\ = (m + n) . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{n + m - 1} + \\dfrac{1}{n + m} \\right)$ <br\/> Khi \u0111\u00f3 ta r\u00fat g\u1ecdn $ \\dfrac{1}{n} + \\dfrac{1}{n + 1} + ... + \\dfrac{1}{n + m - 1} + \\dfrac{1}{n + m}$ \u1edf tr\u00ean t\u1eed v\u00e0 d\u01b0\u1edbi m\u1eabu, b\u00e0i to\u00e1n tr\u1edf n\u00ean d\u1ec5 d\u00e0ng<\/i> <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:54"},{"id":"1692","post_id":"809","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":[[["100"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"R\u00fat g\u1ecdn. <br\/> <br\/> $P = \\dfrac{ \\dfrac{99}{1} + \\dfrac{98}{2} + \\dfrac{97}{3} + ... + \\dfrac{2}{98} + \\dfrac{1}{99} }{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","hint":"","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft t\u1eed s\u1ed1 c\u1ee7a c\u00e1c ph\u00e2n s\u1ed1 $\\dfrac{99}{1}; \\dfrac{99}{2}; .... ; \\dfrac{2}{98}; \\dfrac{2}{99}$ th\u00e0nh hi\u1ec7u c\u1ee7a 100 v\u00e0 m\u1ed9t s\u1ed1 (V\u00ed d\u1ee5 $\\dfrac{99}{1} = \\dfrac{100 - 1}{1}$) <br\/> B\u01b0\u1edbc 2: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 v\u1eeba vi\u1ebft \u1edf b\u01b0\u1edbc 1 th\u00e0nh hi\u1ec7u 2 ph\u00e2n s\u1ed1 (V\u00ed d\u1ee5: $\\dfrac{100 - 1}{1} = \\dfrac{100}{1} - 1$) <br\/> B\u01b0\u1edbc 3: <\/b> \u0110\u1eb7t nh\u00e2n t\u1eed chung ra ngo\u00e0i v\u00e0 r\u00fat g\u1ecdn <\/span> <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} P &= \\dfrac{ \\dfrac{99}{1} + \\dfrac{98}{2} + \\dfrac{97}{3} + ... + \\dfrac{2}{98} + \\dfrac{1}{99} }{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ \\dfrac{100 - 1}{1} + \\dfrac{100 - 2}{2} + \\dfrac{100 - 3}{3} + ... + \\dfrac{100 - 98}{98} + \\dfrac{100 - 99}{99} }{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ \\dfrac{100}{1} - 1 + \\dfrac{100}{2} - 1 + \\dfrac{100}{3} - 1 + ... + \\dfrac{100}{98} - 1 + \\dfrac{100}{99} - 1}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ \\left( \\dfrac{100}{1} + \\dfrac{100}{2} + \\dfrac{100}{3} + ... + \\dfrac{100}{98} + \\dfrac{100}{99} \\right) - \\left( \\underbrace{1 + 1 + ... + 1}_{99 \\hspace{0.3cm} \\text{s\u1ed1 1}} \\right)}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{100 + 100 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{98} + \\dfrac{1}{99} \\right) - 99}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{100 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{98} + \\dfrac{1}{99} \\right) + 100 . \\dfrac{1}{100}}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= \\dfrac{ 100 . \\left( \\dfrac{1}{2} + \\dfrac{1}{3} + ... + \\dfrac{1}{98} + \\dfrac{1}{99} + \\dfrac{1}{100} \\right)}{ \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ... + \\dfrac{1}{99} + \\dfrac{1}{100} } \\\\ &= 100 \\end{align*}$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 100 <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:54"},{"id":"1693","post_id":"809","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":[[["81"],["10"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ednh nhanh. <br\/> <br\/> $A = \\dfrac{1}{2} + \\dfrac{5}{6} + \\dfrac{11}{12} + \\dfrac{19}{20} + \\dfrac{29}{30}+ \\dfrac{41}{42} + \\dfrac{55}{56} + \\dfrac{71}{72} + \\dfrac{89}{90} = \\dfrac{ \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","hint":"Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh hi\u1ec7u c\u1ee7a 1 v\u00e0 ph\u1ea7n b\u00f9 c\u1ee7a n\u00f3","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh hi\u1ec7u c\u1ee7a 1 v\u00e0 ph\u1ea7n b\u00f9 c\u1ee7a n\u00f3 d\u1ea1ng $\\dfrac{n - 1}{n} = 1 - \\dfrac{1}{n}$ <br\/> B\u01b0\u1edbc 2: <\/b> Vi\u1ebft m\u1eabu s\u1ed1 c\u1ee7a m\u1ed7i ph\u00e2n s\u1ed1 tr\u00ean th\u00e0nh t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp <br\/> B\u01b0\u1edbc 3: <\/b> S\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c $\\dfrac{1}{n . (n + 1)} = \\dfrac{1}{n} - \\dfrac{1}{n + 1}$ <br\/> B\u01b0\u1edbc 4: <\/b> T\u00ednh v\u00e0 r\u00fat g\u1ecdn <\/span> <br\/><br\/>B\u00e0i gi\u1ea3i:<\/span><br\/> $\\begin{align*} A &= \\dfrac{1}{2} + \\dfrac{5}{6} + \\dfrac{11}{12} + \\dfrac{19}{20} + \\dfrac{29}{30} + \\dfrac{41}{42} + \\dfrac{55}{56} + \\dfrac{71}{72} + \\dfrac{89}{90} \\\\ &= \\left( 1 - \\dfrac{1}{2} \\right) + \\left( 1 - \\dfrac{1}{6} \\right) + \\left( 1 - \\dfrac{1}{12} \\right) + \\left( 1 - \\dfrac{1}{20} \\right) + \\left( 1 - \\dfrac{1}{30} \\right) + \\left( 1 - \\dfrac{1}{42} \\right) \\\\ & + \\left( 1 - \\dfrac{1}{56} \\right) + \\left( 1 - \\dfrac{1}{72} \\right) + \\left( 1 - \\dfrac{1}{90} \\right) \\\\ &= 9 - \\left( \\dfrac{1}{2} + \\dfrac{1}{6} + \\dfrac{1}{12} + \\dfrac{1}{20} + \\dfrac{1}{30} + \\dfrac{1}{42} + \\dfrac{1}{56} + \\dfrac{1}{72} + \\dfrac{1}{90} \\right) \\\\ &= 9 - \\left( \\dfrac{1}{1 . 2} + \\dfrac{1}{2 . 3} + \\dfrac{1}{3 . 4} + \\dfrac{1}{4 . 5} + \\dfrac{1}{5 . 6} + \\dfrac{1}{6 . 7} + \\dfrac{1}{7 . 8} + \\dfrac{1}{8 . 9} + \\dfrac{1}{9 . 10} \\right) \\\\ &= 9 - \\left( 1 - \\dfrac{1}{2} + \\dfrac{1}{2} - \\dfrac{1}{3} + ... + \\dfrac{1}{9} - \\dfrac{1}{10} \\right) \\\\ &= 9 - \\left( 1 - \\dfrac{1}{10} \\right) \\\\ &= 9 - \\dfrac{9}{10} \\\\ &= \\dfrac{81}{10} \\end{align*}$ <br\/> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 81; 10 <\/span> <br\/> Nh\u1eadn x\u00e9t: V\u1edbi t\u1ed5ng g\u1ed3m nhi\u1ec1u ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{n - 1}{n}$ ta th\u01b0\u1eddng vi\u1ebft m\u1ed7i ph\u00e2n s\u1ed1 \u0111\u00f3 th\u00e0nh hi\u1ec7u c\u1ee7a 1 v\u00e0 ph\u1ea7n b\u00f9 c\u1ee7a n\u00f3 $\\dfrac{n - 1}{n} = 1 - \\dfrac{1}{n}$ <br\/> Kh\u00e9o l\u00e9o vi\u1ebft n th\u00e0nh t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean, bi\u1ebfn \u0111\u1ed1i l\u00e0m xu\u1ea5t hi\u1ec7n c\u00e1c ph\u00e2n s\u1ed1 d\u1ea1ng $\\dfrac{1}{a} - \\dfrac{1}{b} + \\dfrac{1}{b} - \\dfrac{1}{c}$ <br\/> Tri\u1ec7t ti\u00eau c\u00e1c ph\u00e2n s\u1ed1 \u0111\u1ed1i, t\u1eeb \u0111\u00f3 ta d\u1ec5 d\u00e0ng t\u00ednh \u0111\u01b0\u1ee3c gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:14:54"}]}