Question

(x+4)^2 +(x-1)(x+1) =16

 

Answer 1

Ta có :

\(\left(x+4\right)^2+\left(x-1\right)\left(x+1\right)=16\)

\(x^2+8x+16+x^2-1=16\)

\(2x^2+8x-1=0\)

\(\Rightarrow2\left(x^2+4x+4\right)-9=0\)

\(\Rightarrow2\left(x+2\right)^2=9\)

\(\left(x+2\right)^2=4,5\)

\(\Rightarrow\orbr{\begin{cases}x+2=\frac{3\sqrt{2}}{2}\\x+2=-\frac{3\sqrt{2}}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-4+3\sqrt{2}}{2}\\x=-\frac{4+3\sqrt{2}}{2}\end{cases}}\)

Answer 2

(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 ..............