Question

[Đỗ Mạnh Quân]

Tìm số tự nhiên $x$ thỏa mãn: ${\left(2.x+2\right)}^2=64$.

Answer 1

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

`@` `\text {Ans}`

`\downarrow`

\(\left(2\cdot x+2\right)^2=64\)

`\Rightarrow`\(\left(2x+2\right)^2=\left(\pm8\right)^2\)

`\Rightarrow`\(\left[{}\begin{matrix}2x+2=8\\2x+2=-8\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=8+2\\2x=-8+2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=10\\2x=-6\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=10\div2\\x=-6\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

`@` `\text {Kaizuu lv uuu}`