1)

ĐK: \(x\geq 2\)

\(\sqrt{x-2}-3\sqrt{x^2-4}=0\)

\(\Leftrightarrow \sqrt{x-2}-3\sqrt{(x-2)(x+2)}=0\)

\(\Leftrightarrow \sqrt{x-2}(1-3\sqrt{x+2})=0\)

\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}=\frac{1}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=\frac{-17}{9}(\text{loại vì x}\geq 2)\end{matrix}\right.\)

Vậy $x=2$ là nghiệm của pt

2) ĐK: \(x\geq 1\)

Ta có: \(x+\sqrt{x-1}=13\)

\(\Leftrightarrow (x-1)+\sqrt{x-1}+\frac{1}{4}=\frac{49}{4}\)

\(\Leftrightarrow (\sqrt{x-1}+\frac{1}{2})^2=\frac{49}{4}\)

Vì \(\sqrt{x-1}+\frac{1}{2}>0\) nên \(\sqrt{x-1}+\frac{1}{2}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)

\(\Rightarrow \sqrt{x-1}=3\)

\(\Rightarrow x=3^2+1=10\) (thỏa mãn)

Vậy.......

Tên Trung Quốc cơ á

\(\sqrt{2\left(x-3\right)^2+16}\ge4\)

\(\sqrt{4\left(x-3\right)^2+4}\ge2\)

\(\Rightarrow VT\ge6\)

mà \(-x^2+6x-3=-\left(x-3\right)^2+6\le6\)

MÀ VT=VP\(\Rightarrow x=3\)

\(\Leftrightarrow\frac{6x+5}{12x+9}-\frac{3x-7}{12x-9}=\frac{4x^2+10x-7}{16x^2-9}.\)

\(\Leftrightarrow\frac{\left(6x+5\right)\left(12x-9\right)-\left(3x-7\right)\left(12x+9\right)}{\left(3.4.x\right)^2-\left(3.3\right)^2}=\frac{4x^2+10x-7}{16x^2-9}\)

\(\Leftrightarrow\frac{72x^2+6x-45-\left(36x^2-57x-63\right)}{3^2\left(16x^2-9\right)}=\frac{4x^2+10x-7}{16x^2-9}\)

ĐK: \(16x^2-9\ne0\Leftrightarrow x^2\ne\left(\frac{3}{4}\right)^2\Rightarrow x\ne\pm\frac{3}{4}\)

\(\Leftrightarrow72x^2+6x-45-36x^2+57x+63=36x^2+90x-63\)

\(\Leftrightarrow27x=81\Leftrightarrow x=3\)