a) \(\frac{x+1}{x}+1=\frac{3x+1}{x+1}+\frac{1}{x\left(x+1\right)}\)
ĐK: \(x\ne0,x\ne-1\)
\(\Leftrightarrow\frac{x+1}{x}+\frac{x}{x}=\frac{x\left(3x+1\right)+1}{x\left(x+1\right)}\)
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\frac{x\left(3x+1\right)+1}{x\left(x+1\right)}\)
\(\Leftrightarrow2x^2+2x+x+1=3x^2+x+1\)
\(\Leftrightarrow3x^2-2x^2+x-2x-x+1-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Mà \(x\ne0\)
Vậy \(x=2\) là giá trị cần tìm.
b) \(\frac{2x+1}{3}-\frac{3x-2}{2}>\frac{1}{6}\)
\(\frac{2\left(2x+1\right)}{6}-\frac{3\left(3x-2\right)}{6}>\frac{1}{6}\)
\(4x+2-9x+6>1\)
\(-5x+8>1\)
\(-5x>-7\)
\(x< \frac{7}{5}\)
Vậy \(x< \frac{7}{5}\) là giá trị cần tìm.
b, \(\frac{2\left(2x+1\right)}{6}-\frac{3\left(3x-2\right)}{6}>\frac{1}{6}\)
=> 4x+2-9x+6> 1
-5x-7>0
=> x<\(\frac{7}{5}\)
mình không chắc đúng đâu nha !!