\(a.\frac{x-3}{x+7}=\frac{1}{6}\) (đkxđ: \(x+7\neq0\Rightarrow x\neq-7\) )
\(\Rightarrow 6(x - 3) = 1(x + 7)\)
\(\Leftrightarrow 6x - 18 = x + 7\)
\(\Leftrightarrow 6x - x = 7 + 18\)
\(\Leftrightarrow5x=25\Rightarrow x=5\left(TM\right)\)
b. \(\frac{x+4}{x-3}=\frac{x-5}{x+1}\) (đkxđ: \(\begin{cases} x - 3 \neq 0 \\ x + 1 \neq 0 \end{cases} \Leftrightarrow \begin{cases} x \neq 3 \\ x \neq -1 \end{cases}\) )
\(\Rightarrow (x + 4)(x + 1) = (x - 5)(x - 3)\)
\(\Leftrightarrow x^2 + x + 4x + 4 = x^2 - 3x - 5x + 15\)
\(\Leftrightarrow x^2 + 5x + 4 = x^2 - 8x + 15\)
\(\Leftrightarrow x^2 + 5x - x^2 + 8x = 15 - 4\)
\(\Leftrightarrow13x=11\Rightarrow x=\frac{11}{13}\left(TM\right)\)
\(\frac{x-3}{x+7}=\frac{1}{6}\)ĐKXĐ: \(x + 7 \neq 0 \Leftrightarrow x \neq -7\)Ta có:
\(6\cdot (x-3)=1\cdot (x+7)\)
\(6x-18=x+7\)
\(6x-x=7+18\)
\(5x=25\)
\(x=5\text{\ (Tha\ mãn\ ĐKXĐ)}\)Vậy \(x = 5\).Bài 2: Tìm \(x\), biết:
\(\frac{x+4}{x-3}=\frac{x-5}{x+1}\)ĐKXĐ: \(x \neq 3\) và \(x \neq -1\)Ta có:
\((x+4)(x+1)=(x-3)(x-5)\)
\(x^{2}+x+4x+4=x^{2}-5x-3x+15\)
\(x^{2}+5x+4=x^{2}-8x+15\)
\(x^{2}-x^{2}+5x+8x=15-4\)
\(13x=11\)
\(x=\frac{11}{13}\text{\ (Tha\ mãn\ ĐKXĐ)}\)Vậy \(x = \frac{11}{13}\).
