Bài 1:
a) Ta có: \(\left(1-2x\right)\left(1+2x\right)+\left(2x+3\right)^2=34\)
\(\Leftrightarrow1-4x^2+4x^2+12x+9-34=0\)
\(\Leftrightarrow12x-24=0\)
\(\Leftrightarrow12\left(x-2\right)=0\)
Vì 12≠0
nên x-2=0
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x-3\right)^2+\left(3-2x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left[\left(2x-3\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};2\right\}\)
Bài 2:
a) Ta có: \(\frac{2x-5y}{x-y}-\frac{3y}{y-x}\)
\(=\frac{2x-5y}{x-y}+\frac{3y}{x-y}\)
\(=\frac{2x-5y+3y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)
b) Ta có: \(\frac{x^2+3xy}{x^2-9y^2}+\frac{5x-x^2}{x^2-3xy}\)
\(=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}+\frac{x\left(5-x\right)}{x\left(x-3y\right)}\)
\(=\frac{x}{x-3y}+\frac{5-x}{x-3y}\)
\(=\frac{x+5-x}{x-3y}=\frac{5}{x-3y}\)