\(\frac{1}{\sqrt{x-\sqrt{2x+1}}}\)

\(\left|3x+5\right|>x^2-2x+3\)

giải phương trình sau

\(\left|x-2\right|=\left|x^2-4\right|\)

Giải phương trình sau

a, \(\frac{3x}{x^2-x+3}-\frac{2x}{x^2-3x+3}=-1\)

b, \(\frac{1}{\left(x^2+2x+2\right)^2}+\frac{1}{\left(x^2+2x+3\right)^2}=\frac{5}{4}\)

c,\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\)

d,\(\frac{x^2}{2}+\frac{18}{x^2}=13\left(\frac{x}{2}-\frac{3}{x}\right)\)

\(\frac{1}{\left(x^2+2x+2\right)^2}+\frac{1}{\left(x^2-2x+3\right)^2}=\frac{5}{4}\)

\(\left(8x-7\right)\left(8x-5\right)\left(2x-1\right)\left(4x-1\right)=9\)