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1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Dấu " / " là phân số nhé
a) 5/-4 . 16/25 + -5/4 . 9/25
= -5/4 . 16/25 + -5/4 . 9/25
= -5/4 . ( 16/25 + 9/25 )
= -5/4 . 1
= -5/4
b) 4 11/23 - 9/14 + 2 12/23 - 5/4
= 103/23 - 9/14 + 58/23 - 5/4
= 103/23 + 58/23 - 9/14 - 5/4
= 7 - 9/14 - 5/4
= 143/28
c) 2 13/27 - 7/15 + 3 14/27 - 8/15
= 67/27 - 7/15 + 95/27 - 8/15
= 67/27 + 95/27 - 7/15 - 8/15
= 6 - 7/15 - 8/15
= 5
\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\\ A=15.\left(\dfrac{9}{15}-\dfrac{10}{15}\right)+1\\ A=15.\dfrac{-1}{15}+1\\ A=-1+1\\ A=0\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{9}.\dfrac{9}{11}+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.1+\dfrac{12}{7}\\ C=\dfrac{-5}{7}+\dfrac{12}{7}\\ C=1\)
3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
\(\Leftrightarrow2x-6-3+6x=4+4-4x\)
\(\Leftrightarrow8x-9=8-4x\)
\(\Leftrightarrow8x+4x=8+9\)
\(\Leftrightarrow12x=17\)
\(\Leftrightarrow x=\dfrac{17}{12}\)
Vậy \(x=\dfrac{17}{12}\)
4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)
\(\Leftrightarrow6x-12-4-4x=12-9x-12\)
\(\Leftrightarrow6x-4-4x=12-9x\)
\(\Leftrightarrow2x-4=12-9x\)
\(\Leftrightarrow2x+9x=12+4\)
\(\Leftrightarrow11x=16\)
\(\Leftrightarrow x=\dfrac{16}{11}\)
Vậy \(x=\dfrac{16}{11}\)
\(y+30\%y=-1,3\\ 130\%y=-1,3\\ \Rightarrow y=\dfrac{-1,3}{130\%}=-1\)
\(x:\dfrac{4}{28}=\dfrac{13}{-19}+\dfrac{8}{25}\\ 7x=-\dfrac{173}{475}\\ x=-\dfrac{\dfrac{173}{475}}{7}=-\dfrac{173}{3325}\)
\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)
\(\Rightarrow\dfrac{M}{2}=\dfrac{6:2}{2.5}+...+\dfrac{6:2}{47.50}\)
\(=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}\)
\(=\dfrac{12}{25}\)
\(\Rightarrow M=\dfrac{12}{25}.2=\dfrac{24}{25}\)
\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)
\(\Rightarrow2K=\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{43.45}\)
\(=\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\)
\(=\dfrac{1}{9}-\dfrac{1}{45}\)
\(=\dfrac{4}{45}\)
\(\Rightarrow K=\dfrac{4}{45}:2=\dfrac{2}{45}\)
\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)
\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{5}+\dfrac{6}{5}-\dfrac{6}{8}+\dfrac{6}{8}-\dfrac{6}{11}+...+\dfrac{6}{47}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\left(\dfrac{150}{50}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\dfrac{144}{50}\)
\(M=\dfrac{144}{25}\)
\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)
\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\left(\dfrac{5}{45}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\dfrac{4}{45}\)
\(K=\dfrac{2}{45}\)
\(D=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)
Áp dụng công thức tổng quát \(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
Ta có:
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\\ D=\dfrac{1}{3}-\dfrac{1}{87}\\ D=\dfrac{28}{87}\)
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