\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)
\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)
\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)
\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)
\(\Leftrightarrow3x^2-7x+2=0\)
\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)
Tự cho đkxđ nha!!!
<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)
<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)
<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)
<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)
=> 2x = 0
<=> x = 0 (TM)
Vậy ...
\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)
\(\left(x+1\right)\left(3x-2\right)\left(2-3x\right)-x\left(3x-2\right)\left(2-3x\right)=7x\left(2-3x\right)+2\left(x+1\right)\left(3x-2\right)\)
\(-9x^2+12x-4=16x-15x^2-4\)
\(-9x^2+12x=16x-15x^2\)
\(9x^2-12x+16x-15x^2=0\)
\(-6x^2+4x=0\)
\(-2x\left(3x-2\right)=0\)
\(Th1:-2x=0\Leftrightarrow x=0\)
\(Th2:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\left(x\ne-1;x\ne\frac{2}{3}\right)\)
\(\Leftrightarrow\frac{x+1-x}{x+1}-\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{3x-2}=0\)
\(\Leftrightarrow\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}-\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2x+2}{\left(3x-2\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{3x-2-7x+2x+2}{\left(3x-2\right)\left(x+1\right)}=0\)
\(\Rightarrow-2x=0\)
<=> x=0 (tm)
Vậy x=0
