Question

[bach nhac lam]

gpt : a) \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

b) \(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)

c) \(\left(x-2\right)\left(x+2\right)+4\left(x-2\right)\sqrt{\frac{x+2}{x-2}}=-3\)

@Akai Haruma, @Bonking, @Nguyễn Việt Lâm

Answer 1

[Trần Thanh Phương]

c) Đặt \(\left\{{}\begin{matrix}x-2=a\\x+2=b\end{matrix}\right.\)

\(pt\Leftrightarrow ab+4a\cdot\sqrt{\frac{b}{a}}=-3\)

\(\Leftrightarrow ab+\sqrt{\frac{16a^2\cdot b}{a}}+3=0\)

\(\Leftrightarrow ab+\sqrt{16ab}+3=0\)

\(\Leftrightarrow ab+4\sqrt{ab}+3=0\)

\(\Leftrightarrow ab+\sqrt{ab}+3\sqrt{ab}+3=0\)

\(\Leftrightarrow\sqrt{ab}\left(\sqrt{ab}+1\right)+3\left(\sqrt{ab}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}+1\right)=0\)

Dễ thấy \(VT>0\forall x\)

Do đó pt vô nghiệm

Answer 2

[Akai Haruma]

b)

Đặt \(\left\{\begin{matrix} \sqrt[3]{7-x}=a\\ \sqrt[3]{x-5}=b\end{matrix}\right.\). PT đã cho trở thành:

\(\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\)

\(\Leftrightarrow (a-b)\left(\frac{1}{a+b}-\frac{a^2+ab+b^2}{2}\right)=0\)

Nếu \(a-b=0\Leftrightarrow a=b\Leftrightarrow a^3=b^3\Leftrightarrow 7-x=x-5\)

\(\Leftrightarrow x=6\) (thỏa mãn)

Nếu \(\frac{1}{a+b}-\frac{a^2+ab+b^2}{2}=0\)

\(\Leftrightarrow (a^2+ab+b^2)(a+b)=2=a^3+b^3\)

\(\Leftrightarrow a^2b+ab^2=0\Leftrightarrow ab(a+b)=0\)

Hiển nhiên $a+b\neq 0$ (để biểu thức có nghĩa)

Do đó \(\left[\begin{matrix} a=0\\ b=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=7\\ x=5\end{matrix}\right.\)

Vậy........