Question

Giai các ptr sau

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

b,\(\dfrac{8800}{x-2}-\dfrac{8800}{x}=20\)

c,\(\dfrac{1}{x}+\dfrac{1}{x+5}=\dfrac{1}{6}\)

d,\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

Answer 1

a: \(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)

=>-2x^2+x+1-2x^2+2x=0

=>-4x^2+3x+1=0

=>4x^2-3x-1=0

=>4x^2-4x+x-1=0

=>(x-1)(4x+1)=0

=>x=1(loại) hoặc x=-1/4(nhận)

b: \(\Leftrightarrow\dfrac{440}{x-2}-\dfrac{440}{x}=1\)

=>x(x-2)=440x-440x+880

=>x^2-2x-880=0

=>\(x=1\pm\sqrt{881}\)

c: \(\Leftrightarrow\dfrac{x+5+x}{x\left(x+5\right)}=\dfrac{1}{6}\)

=>x^2+5x=6(2x+5)

=>x^2+5x-12x-30=0

=>x^2-7x-30=0

=>(x-10)(x+3)=0

=>x=10 hoặc x=-3

d: =>(x-1)(x+1)-x=2x-1

=>x^2-1-x=2x-1

=>x^2-x-2x=0

=>x(x-3)=0

=>x=0(loại) hoặc x=3(nhận)