Question

Giải pt:

a,\(\dfrac{x+1}{x-1}-\dfrac{2}{x\left(x-2\right)}=\dfrac{1}{x}\)

b, \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

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

d, \(\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}+\dfrac{4}{x^2+2x-3}=0\)

Answer 1

d) \(\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}+\dfrac{4}{x^2+2x-3}=0\) (ĐKXĐ: \(x\ne1;-3\))

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)+4}{\left(x+3\right)\left(x-1\right)}=0\)

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

\(\Leftrightarrow x^2+4x+3-x^2-x+2+4=0\)

\(\Leftrightarrow3x+9=0\Leftrightarrow x=-3\left(loại\right)\)

vậy phương trình đã cho vô nghiệm

Answer 2

c)\(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\) (ĐKXĐ: \(x\ne1\))

\(\Leftrightarrow\dfrac{2\left(x^2+x+1\right)-3x^2}{x^3-1}=\dfrac{x\left(x-1\right)}{x^3-1}\)

\(\Rightarrow2x^2+2x+2-3x^2=x^2-x\)

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

\(\left(x-2\right)\left(-2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\-2x-1=0\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)

vậy tập nghiệm của phương trình là S={2;-0,5)

Answer 3

b)\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)};ĐKXĐ:x\ne\left\{0;2\right\}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

=>x²+2x-x+2=2<=>x²+x=2-2<=>x(x+1)=0

=>\(\left\{{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy tập nghiệm của PT là S={-1}