giải phương trình
a: \(\frac{x-6}{x-4}\) = \(\frac{x}{x-2}\)
b:1 + \(\frac{2x-5}{x-2}\) - \(\frac{3x-5}{x-1}\) = 0
c: \(\frac{x-3}{x-2}\) - \(\frac{x-2}{x-4}\) = 3\(\frac{1}{5}\)
d:\(\frac{x+1}{x-2}\) - \(\frac{x-1}{x+2}\) = \(\frac{2\left[x^2+2\right]}{x^2-4}\)
e:\(\frac{4}{x^2+2x-3}\) = \(\frac{2x-5}{x+3}\) - \(\frac{2x}{x-1}\)
j: \(\frac{3}{x^2+x-2}\) - \(\frac{1}{x-1}\) = \(\frac{-7}{x+2}\)
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)
\(d.\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\\ \Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)-2\left(x^2+2\right)=0\\ \Leftrightarrow x^2+2x+x+2-x^2+2x+x-2-2x^2-4=0\\ \Leftrightarrow-2x^2+6x-4=0\\ \Leftrightarrow-2x^2+4x+2x-4=0\\ \Leftrightarrow\left(-2x^2+4x\right)+\left(2x-4\right)=0\\ \Leftrightarrow-2x\left(x-2\right)+2\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{2}{2}=1\end{matrix}\right.\)
Câu e mik chịu!! ^ _ ^
