Question

[Phương Trinh]

 

giúp mik nha

Answer 1

[Nguyễn Minh Đăng]

a) đk: \(x\ne\left\{1;2\right\}\)

Ta có: \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=0\)

\(\Rightarrow x^2-3x+2+2x^2-7x+5-3x^2+11x-10=0\)

\(\Leftrightarrow x-3=0\Rightarrow x=3\)

Vậy x = 3

b) đk: \(x\ne\left\{0;2\right\}\)

Ta có: \(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)

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

\(\Rightarrow x^2+2x-2=x-2\)

\(\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\) vì x khác 0 nên

=> \(x+1=0\Rightarrow x=-1\)

Vậy x = -1

Answer 2

[Nguyễn Minh Đăng]

c) đk: \(x\ne\pm3\)

Ta có: \(\frac{x+2}{x-3}+\frac{x-2}{x+3}-\frac{2\left(x^2+6\right)}{x^2-9}=0\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x+3\right)+\left(x-2\right)\left(x-3\right)-2\left(x^2+6\right)}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow x^2+5x+6+x^2-5x+6-2x^2-12=0\)

\(\Leftrightarrow0x=0\) (luôn đúng)

Vậy mọi x thực thỏa mãn đk thì PT luôn có nghiệm

d) đk: \(x\ne-1\)

Ta có: \(\frac{-7x^2+4}{x^3+1}=\frac{5}{x^2-x+1}-\frac{1}{x+1}\)

\(\Leftrightarrow\frac{-7x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{5\left(x+1\right)-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow-7x^2+4=5x+5-x^2+x-1\)

\(\Leftrightarrow6x^2+6x=0\)

\(\Leftrightarrow6x\left(x+1\right)=0\) vì x + 1 khác 0

=> x = 0

Vậy x = 0