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\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\) (đpcm)


\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}=\frac{2}{11}\)

\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{121}-1\right)=-\frac{3}{4}.-\frac{8}{9}....-\frac{120}{121}=\frac{-1.3}{2.2}\cdot\frac{-2.4}{3.3}....-\frac{10.12}{11.11}=\frac{-1}{2}.-\frac{12}{11}=\frac{6}{11}\)
a) |2x + 5| = 3 - x
\(\Rightarrow\orbr{\begin{cases}3-x=2x+5\\3-x=-\left(2x+5\right)\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3-5=2x+x\\3-x=-2x-5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=-2\\-x+2x=-5-3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=-8\end{cases}}\)
Vậy \(x\in\left\{\frac{-2}{3};-8\right\}\)
b) lm tương tự câu a