Tìm x:
a/ \(\left(3x-5\right)^2-3x+5=0\)
b/ \(3x^2+8x-3=0\)
c/ \(\left(x+1\right)^2=\left(2x-5\right)^2\)
Hướng dẫn giải
a/ \(\left(3x-5\right)^2-3x+5=0\)
\(\Leftrightarrow\left(3x-5\right)^2-\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x-5-1\right)=0\Leftrightarrow\left(3x-5\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}\)
b/ \(3x^2+8x-3=0\Leftrightarrow3x^2+9x-x-3=0\)
\(3x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
c/ \(\left(x+1\right)^2=\left(2x-5\right)^2\Leftrightarrow\left(x+1\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(x+1+2x-5\right)\left(x+1-2x+5\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(6-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=6\end{cases}}\)