Question

Answer 1

a: ĐKXD: x<>-3; x<>1

\(P=\dfrac{x^2-1+x^2+3x-x^2+3x+10}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+6x+9}{\left(x+3\right)\left(x-1\right)}=\dfrac{x+3}{x-1}\)

b: |5x-2|=x+2

TH1: x>=2/5

=>5x-2=x+2

=>4x=4

=>x=1(loại)

TH2: x<2/5

=>-5x+2=x+2

=>-6x=0

=>x=0(nhận)

Khi x=0 thì \(P=\dfrac{0+3}{0-1}=-3\)

c: 2P=x-1

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

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

=>x^2-2x+1=2x+6

=>x^2-4x-5=0

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1