\(\left(x+3\right)^2-\left(4-x\right)\left(4+x\right)=10\)
<=> \(x^2+6x+9-\left(16-x^2\right)=10\)
<=> \(2x^2+6x-17=0\)
<=> \(x^2+3x-\frac{17}{2}=0\)
<=> \(\left(x+\frac{3}{2}\right)^2-\frac{43}{4}=0\)
<=> \(\left(x+\frac{3}{2}+\frac{\sqrt{43}}{2}\right)\left(x+\frac{3}{2}-\frac{\sqrt{43}}{2}\right)=0\)
<=> \(\orbr{\begin{cases}x+\frac{3}{2}+\frac{\sqrt{43}}{2}=0\\x+\frac{3}{2}-\frac{\sqrt{43}}{2}=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{-3-\sqrt{43}}{2}\\x=\frac{\sqrt{43}-3}{2}\end{cases}}\)
Vậy...
\((x+3)^2-(4-x)(4+x)=10\)
\(\Rightarrow x^2+6x+9-(16+4x-4x+x^2)=10\)
\(\Rightarrow x^2+6x+9-16-x^2=10\)
\(\Rightarrow6x+9=26\)
\(\Rightarrow6x=17\)
\(\Rightarrow x\in\varnothing\)
