Question

X/x+1-2x-3/x-1=2x+3/x^2-1 giải phương trình chứa ẩn ở mẫu!!😀😀

Answer 1

\(\frac{x}{x+1}-\frac{2x-3}{x-1}=\frac{2x+3}{x^2-1}\) ĐKXĐ: x ≠ 1; x ≠ -1

⇔x(x - 1) - (2x - 3)(x + 1) = 2x + 3

⇔ x² - x - 2x² + 3x - 2x + 3 = 2x + 3

⇔ -x² - 2x = 0

⇔ -x(x + 2) = 0

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

Vậy nghiệm của pt là x = 0; x = -2

Answer 2

ĐKXĐ: x≠1; x≠-1

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

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

\(\Leftrightarrow x^2-x-\left(2x^2+2x-3x-3\right)-\left(2x+3\right)=0\)

\(\Leftrightarrow x^2-x-2x^2+x+3-2x-3=0\)

\(\Leftrightarrow-x^2-2x=0\)

\(\Leftrightarrow-x\left(x+2\right)=0\)

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

Vậy: x∈{0;-2}