giúp mình với

ai giúp mình với

tính phép tính trên nha mọi người

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}\)

\(=\frac{1}{9900}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)=\frac{1}{9900}-\frac{98}{99}=\frac{-9799}{9900}\)

ta có :

a) \(\left(-\frac{2}{3}\right)^2:\frac{1}{3}-\left|-1\frac{1}{2}\right|=\frac{4}{9}:\frac{1}{3}-\frac{3}{2}=\frac{4}{3}-\frac{3}{2}=-\frac{1}{6}\)

b) \(\left(\frac{1}{2}-\frac{3}{5}\right)^2+\frac{2}{3}\left|\frac{3}{4}-\frac{1}{2}\right|+2012^0=\left(-\frac{1}{10}\right)^2+\frac{2}{3},\frac{1}{4}+2012^0\)

\(=\frac{1}{100}+\frac{1}{6}+1=\frac{353}{300}\)

c) \(\left(3^2:\frac{1}{3}\right)+2^3+\frac{1}{2}+\frac{1}{4}-6=3^3+2^3+\frac{3}{4}-6=29\frac{3}{4}\)

\(3\left(2x+1\right)^2-5=22\)

\(\Rightarrow3\left(2x+1\right)^2=27\)

\(\Rightarrow\left(2x+1\right)^2=9\)

\(\Rightarrow\left(2x+1\right)^2-3^2=0\)

\(\Rightarrow\left(2x+1-3\right)\left(2x+1+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-2=0\\2x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)