Question

[ngoc bich 2]

Tìm x 

(x+3)^2 - (4-x)(4+x)=10

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Answer 1

[Không Tên]

\(\left(x+3\right)^2-\left(4-x\right)\left(4+x\right)=10\)

<=>   \(x^2+6x+9-\left(16-x^2\right)=10\)

<=>   \(2x^2+6x-17=0\)

<=>  \(x^2+3x-\frac{17}{2}=0\)

<=>  \(\left(x+\frac{3}{2}\right)^2-\frac{43}{4}=0\)

<=>  \(\left(x+\frac{3}{2}+\frac{\sqrt{43}}{2}\right)\left(x+\frac{3}{2}-\frac{\sqrt{43}}{2}\right)=0\)

<=>  \(\orbr{\begin{cases}x+\frac{3}{2}+\frac{\sqrt{43}}{2}=0\\x+\frac{3}{2}-\frac{\sqrt{43}}{2}=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=\frac{-3-\sqrt{43}}{2}\\x=\frac{\sqrt{43}-3}{2}\end{cases}}\)

Vậy...

Answer 2

[phạm văn tuấn]

\((x+3)^2-(4-x)(4+x)=10\)

\(\Rightarrow x^2+6x+9-(16+4x-4x+x^2)=10\)

\(\Rightarrow x^2+6x+9-16-x^2=10\)

\(\Rightarrow6x+9=26\)

\(\Rightarrow6x=17\)

\(\Rightarrow x\in\varnothing\)