Đáp án:
Giải thích các bước giải:
Bài 1
a $\frac{3}{x-4}$ $+$ $\frac{x}{x+4}=2$
⇔ĐKXĐ \(\left[ \begin{array}{l}x-4∦0\\x+4∦0\end{array} \right.\)
⇔ $\left \{ {{x∦4} \atop {x∦-4}} \right.$
ĐKXĐ x∦±4
b A=$(\frac{x}{x^{2}-3x}+$ $\frac{1}{x+3})$ $(1-\frac{3}{x})$
A= $[\frac{x}{x(x-3)}$ $+\frac{1}{x+3}]$$(\frac{x-3}{x})$
A= $[\frac{x(x+3)+x(x-3)}{x(x^{2}-9)}$ ] $(\frac{x-3}{x})$
A= $[\frac{x(x+3+x-3)}{x(x^{2}-9)}]$$(\frac{x-3}{x})$
A= $[\frac{2x^{2}}{x(x+3)(x-3)}]$ $(\frac{x-3}{x})$
A=$\frac{2}{x+3}$
Vậy A=$\frac{2}{x+3}$
Bài 4
a 2x+6=0 b 7-2x=22-3x
⇔2x=-6 ⇔-2x-3x=22-7
⇔x=-3 ⇔-5x=15
Vậy S={-3} ⇔x=-3. Vậy S={-5}
c 5-(x-6)=4(3-2x)
⇔5-x+6=12-8x
⇔-x+8x=12-5-6
⇔7x=1
⇔x=$\frac{1}{7}$
Vậy S={$\frac{1}{7}$ }
d x(x-2)+x(4-x)=1
⇔x²-2x+4x-x²=1
⇔2x=1
⇔x=$\frac{1}{2}$
Vậy S={$\frac{1}{2}$}
Bài 5
a (2x-3)(2-x)=0
⇔\(\left[ \begin{array}{l}2x=0\\2-x=0\end{array} \right.\)
⇔$\left \{ {{x=\frac{3}{2} } \atop {x=2}} \right.$
Vậy S={$\frac{3}{2},2$ }
b (3-6x)(x²+1)=0
⇔3-6x=0
⇔-6x=-3
⇔x=$\frac{1}{2}$
Vậy S={$\frac{1}{2}$ }
