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Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+..........+\frac{1}{8}.\frac{1}{9}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{8.9}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+....+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{10.11}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
\(\text{c,d cơ bản tự làm nha }\)
A=>1.1/2.3+1.1/3.4+1.1/4.5+1.1/5.6+1.11/6.7+.1/7.8+1.1/8.9
=>1/2.3+1/3.4+1/4.5+1/6.7+1/7.8+1/8.9
=>1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9
=>1/2-1/9=>9/18-2/18=>7/18
Vậy A= 7/18