1) Ta có: \(\left(x^2+2x-5\right)^2=\left(x^2-x+5\right)^2.\)
<=> \(\left(x^2+2x-5\right)^2-\left(x^2-x+5\right)^2=0\)
<=> \(\left(3x-10\right)\left(2x^2+x\right)=0\)
<=> \(\left(3x-10\right)\cdot x\cdot\left(2x+1\right)=0\)
TH1: 3x-10=0 <=> x=10/3
TH2: x=0
TH3: 2x+1=0 <=> x=-1/2
2) Ta có: \(\left(x-5\right)\left(x-6\right)\left(x+2\right)\left(x+3\right)=180\)
<=> \(\left(x-5\right)\left(x+2\right)\cdot\left(x-6\right)\left(x+3\right)=180\)
<=> \(\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\)
Đặt t = \(x^2-3x-14\)
Ta được pt <=> \(\left(t-4\right)\left(t+4\right)=180\)
<=> \(t^2-16=180\)
<=> \(t^2=196\)<=> \(\orbr{\begin{cases}t=14\\t=-14\end{cases}}\)
TH1: t=14 <=> \(x^2-3x-14=14\)
<=> \(x^2-3x-28=0\)
<=> \(\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)
TH2: t=-14 <=> \(x^2-3x-14=-14\)
<=> \(x\left(x-3\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
