Bài 2:

a: 5x+2=2x-7

=>5x-2x=-7-2

=>3x=-9

=>x=-3

b:

ĐKXĐ: x∉{0;-1}

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

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

=>\(3x^2+4x+1=x^2+3x-\left(x^2-1\right)=3x+1\)

=>\(3x^2+x=0\)

=>x(3x+1)=0

=>x=0(loại) hoặc x=-1/3(nhận)

c: ĐKXĐ: x∉{2;-1}

\(\frac{2}{x+1}+\frac{1}{2-x}=\frac{3x-11}{x^2-x-2}\)

=>\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x-2\right)\left(x+1\right)}\)

=>\(\frac{2\left(x-2\right)-\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{3x-11}{\left(x-2\right)\left(x+1\right)}\)

=>2(x-2)-(x+1)=3x-11

=>3x-11=2x-4-x-1=x-5

=>3x-x=-5+11

=>2x=6

=>x=3(nhận)

Bài 3:

a: ĐKXĐ: x∉{5;-5}

\(\frac{x+5}{x-5}+\frac{20}{25-x^2}=\frac{x-5}{x+5}\)

=>\(\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{20}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}\)

=>\(\left(x+5\right)^2-20=\left(x-5\right)^2\)

=>\(x^2+10x+25-20=x^2-10x+25\)

=>10x+5=-10x+25

=>20x=20

=>x=1(nhận)

b: \(\left(x-2\right)^4-3\left(x^2-4x+5\right)=1\)

=>\(\left(x-2\right)^4-3\left(x^2-4x+4+1\right)-1=0\)

=>\(\left(x-2\right)^4-3\left(x-2\right)^2-4=0\)

=>\(\left\lbrack\left(x-2\right)^2-4\right\rbrack\left\lbrack\left(x-2\right)^2+1\right\rbrack=0\)

=>\(\left(x-2\right)^2-4=0\)

=>(x-2-2)(x-2+2)=0

=>x(x-4)=0

=>x=0 hoặc x=4