Ta có :
\(\left(x+4\right)^2+\left(x-1\right)\left(x+1\right)=16\)
\(x^2+8x+16+x^2-1=16\)
\(2x^2+8x-1=0\)
\(\Rightarrow2\left(x^2+4x+4\right)-9=0\)
\(\Rightarrow2\left(x+2\right)^2=9\)
\(\left(x+2\right)^2=4,5\)
\(\Rightarrow\orbr{\begin{cases}x+2=\frac{3\sqrt{2}}{2}\\x+2=-\frac{3\sqrt{2}}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-4+3\sqrt{2}}{2}\\x=-\frac{4+3\sqrt{2}}{2}\end{cases}}\)
(x+4)2+(x-1)(x+1)=16
(x+4)2+x2-1=16
x2+8x+16+x2-1=16
2x2+8x-1=0
=>2(x2+4x+4)-9=0
=>2(x+2)2=9
(x+2)2=4,5
............hết bt lm ròi ..............