\(\dfrac{x-1}{x-2}=\dfrac{x+2}{x+3}\)
\(\Leftrightarrow1+\dfrac{1}{x-2}=1+\dfrac{-1}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x-2}=\dfrac{-1}{x+3}\)
\(\Leftrightarrow x+3=-x+2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{2}\)
=> (x-1)(x+3) = (x-2)2
=>x2 + 2x - 3 = x2 - 2x + 4
=>x2 + 2x - x2 + 2x = 4+3
=>4x = 7
=> x = 7/4
Ta có : \(\dfrac{x-1}{x-2}=\dfrac{x-2}{x+3}\)
Nên : \(\left(x-1\right).\left(x+3\right)=\left(x-2\right).\left(x-2\right)\)
Hay :\(x^2+3x-x-3=\left(x-2\right)^2\)
\(\Leftrightarrow x^2+2x-3=x^2-4x+4\)
\(\Leftrightarrow x^2-x^2+2x+4x=3+4\)
\(\Leftrightarrow6x=7\)
\(\Rightarrow x=\dfrac{7}{6}\) .
Vây \(x=\dfrac{7}{6}\)
\(\dfrac{x-1}{x-2}=\dfrac{x-2}{x+3}\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)^2\)
\(\Leftrightarrow x^2+2x-3=x^2-4x+4\)
\(\Leftrightarrow x^2-x^2+2x+4x=4+3\)
\(\Leftrightarrow6x=7\Leftrightarrow x=\dfrac{7}{6}\)