Question

[Võ Lan Nhi]

Giải các pt sau:

a) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\)

b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)

Answer 1

[Võ Đông Anh Tuấn]

a ) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\) (1)

ĐKXĐ : x\(\ne1;-2.\)

\(\left(1\right)\Leftrightarrow x+2-7x+7=3\)

\(\Leftrightarrow-6x=-6\)

\(\Leftrightarrow x=1\left(loại\right)\)

Vậy pt vô nghiệm .

b ) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)

Đặt \(x^2+2x+1=t\) ta được :

\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow6t^2+12t+6t^2+12t+6=7\left(t^2+3t+2\right)\)

\(\Leftrightarrow5t^2+3t-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{8}{5}\end{matrix}\right.\)

Khi t = 1

\(\Leftrightarrow\left(x+1\right)^2=1\)

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

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

Khi \(t=-\dfrac{8}{5}\)

\(\Leftrightarrow\left(x+1\right)^2=-\dfrac{8}{5}\) ( vô lí )

Vậy ............