a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 7+\sqrt{2x}=(3+\sqrt{5})^2=14+6\sqrt{5}$
$\Leftrightarrow \sqrt{2x}=7+6\sqrt{5}$
$\Leftrightarrow 2x=(7+6\sqrt{5})^2=229+84\sqrt{5}$
$\Leftrightarrow x=114,5+42\sqrt{5}$ (tm)
b. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow x^2-6x+9=4+2\sqrt{3}$
$\Leftrightarrow (x-3)^2=(\sqrt{3}+1)^2$
$\Leftrightarrow (x-3-\sqrt{3}-1)(x-3+\sqrt{3}+1)=0$
$\Leftrightarrow (x-4-\sqrt{3})(x-2+\sqrt{3})=0$
$\Leftrightarrow x-4-\sqrt{3}=0$ hoặc $x-2+\sqrt{3}=0$
$\Leftrightarrow x=4+\sqrt{3}$ hoặc $x=2-\sqrt{3}$
c. ĐKXĐ: $x\geq \frac{4}{3}$ hoặc $x\leq 0$
PT \(\Rightarrow \left\{\begin{matrix}
2x-3\geq 0\\
3x^2-4x=(2x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\geq \frac{3}{2}\\
x^2-8x+9=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{3}{2}\\ (x-4-\sqrt{7})(x-4+\sqrt{7})=0\end{matrix}\right.\Rightarrow x=4+\sqrt{7}\)
d. ĐKXĐ: $5\leq x\leq 7$
Đặt $\sqrt{7-x}=a; \sqrt{x-5}=b(a,b\geq 0)$ thì pt trở thành:
\(\left\{\begin{matrix}
a^2+b^2=2\\
\frac{a^3+b^3}{a+b}=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
a^2+b^2=2\\
a^2-ab+b^2=2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a^2+b^2=2\\ ab=0\end{matrix}\right.\Rightarrow (a,b)=(0,\sqrt{2}), (\sqrt{2}, 0)\) (do $a,b\geq 0$)
Với $(a,b)=(0,\sqrt{2})\Rightarrow x=7$ (tm)
Với $(a,b)=(\sqrt{2},0)\Rightarrow x=5$ (tm)
