Bài 1:
a: \(3x-10=2\left(x-\frac12\right)\)
=>3x-10=2x-1
=>3x-2x=-1+10
=>x=9
b: ĐKXĐ: x<>0; x<>1
\(\frac{x-2}{x-1}-\frac{1}{x}=\frac{x^2+2}{2x^2-2x}\)
=>\(\frac{x\left(x-2\right)-\left(x-1\right)}{x\left(x-1\right)}=\frac{x^2+2}{2x\left(x-1\right)}\)
=>\(\frac{x^2-2x-x+1}{x\left(x-1\right)}=\frac{x^2+2}{2x\left(x-1\right)}\)
=>\(\frac{2\left(x^2-3x+1\right)}{2x\left(x-1\right)}=\frac{x^2+2}{2x\left(x-1\right)}\)
=>\(2\left(x^2-3x+1\right)=x^2+2\)
=>\(2x^2-6x+2=x^2+2\)
=>\(x^2-6x=0\)
=>x(x-6)=0
=>x=0(loại) hoặc x=6(nhận)
c:|2x-3|=-x-2
=>\(\begin{cases}-x-2\ge0\\ \left(2x-3\right)^2=\left(-x-2\right)^2=\left(x+2\right)^2\end{cases}\)
=>\(\begin{cases}x\le-2\\ \left(2x-3-x-2\right)\left(2x-3+x+2\right)=0\end{cases}\Rightarrow\begin{cases}x\le-2\\ \left(x-5\right)\left(3x-1\right)=0\end{cases}\)
=>x∈∅
Bài 2:
a: 2-5x<-2x-7
=>-5x+2x<-7-2
=>-3x<-9
=>3x>9
=>x>3
b: \(\frac{1-2x}{4}-1\ge\frac{1-5x}{8}\)
=>\(\frac{2\left(1-2x\right)}{8}-\frac88\ge\frac{1-5x}{8}\)
=>2(1-2x)-8>=1-5x
=>2-4x-8>=1-5x
=>-4x-4>=1-5x
=>-4x+5x>=1+4
=>x>=5
Bài 3:
a: ĐKXĐ: x∉{1;-1;-1/2}
\(P=\left(\frac{x-1}{x+1}-\frac{x}{x-1}-\frac{3x+1}{1-x^2}\right):\frac{2x+1}{x^2-1}\)
\(=\frac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(=\frac{x^2-2x+1-x^2-x+3x+1}{2x+1}=\frac{2}{2x+1}\)
b: \(P=\frac{3}{x-1}\)
=>\(\frac{2}{2x+1}=\frac{3}{x-1}\)
=>3(2x+1)=2(x-1)
=>6x+3=2x-2
=>6x-2x=-3-2
=>4x=-5
=>x=-5/4(nhận)
