Đáp án:1)$x=\frac{41}{4}$
3)$x=\frac{-8}{3}$
5)x=5
7)\(\left[ \begin{array}{l}x=-1\\x=-\frac{4}{3}\end{array} \right.\)
9)\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Giải thích các bước giải:
1)$\frac{7}{x+2}=\frac{3}{x-5} (đk: x\neq-2; x\neq5)$
⇔$\frac{7}{x+2}-\frac{3}{x-5}=0$
⇒ $7(x-5)-3(x+2)=0⇔ 7x-35-3x-6=0$
⇔$ 4x=41⇔ x=\frac{41}{4}$
3)$\frac{x^{2}-4}{x}=\frac{2x+3}{2} (đk: x\neq0)$
⇔$\frac{x^{2}-4}{x}-\frac{2x+3}{2}=0$
⇒$2(x^{2}-4)=x(2x+3)⇔ 2x^{2}-8=2x^{2}+3x$
⇔$3x=-8⇔ x=\frac{-8}{3}$
5)$\frac{2-x}{x-1}+\frac{x-3}{x+1}=\frac{2x}{1-x^{2}} (đk: x\neq±1)$
⇔$\frac{2-x}{x-1}+\frac{x-3}{x+1}+\frac{2x}{1-x^{2}} =0$
⇒ $(2-x)(x+1)+(x-3)(x-1)+2x=0⇔ -x^{2}+x+2+x^{2}-4x+3+2x=0$
⇔$ x=5$
7)$\frac{2x+1}{x-1}=\frac{5(x-1)}{x+1} (đk: x\neq±1)$
⇔$\frac{2x+1}{x-1}-\frac{5(x-1)}{x+1}=0$
⇒$(2x+1)(x+1)-5(x-1)(x-1)=0$
⇔$2x^{2}+3x+1-5(x^{2}+2x+1)=0$
⇔$2x^{2}+3x+1-5x^{2}-10x-5=0$
⇔$-3x^{2}-7x-4=0 ⇔ (x+1)(x+\frac{4}{3})=0$
⇔\(\left[ \begin{array}{l}x+1=0\\x+\frac{4}{3}=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=-1\\x=-\frac{4}{3}\end{array} \right.\)
9)$\frac{x}{x-1}-\frac{2x}{x^{2}-1}=0 (đk: x\neq±1)$
⇒$ x(x+1)-2x=0⇔ x^{2}+x-2x=0⇔ x^{2}-x=0$
⇔ $x(x-1)=0$⇔ \(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
