Câu 1: (3,0 điểm) Giải các phương trình sau: a) $8x-3=5x+7$ b)$(4x-1)(x+2) = 0$ c) $\frac{2x-1}{3}$ +$\frac{x+4}{2}$ =$\frac{5x+20}{6}$  d) $\frac{3}{x-2}$ +$\frac{1}{x+2}$ =$\frac{5x-3}{x^{2}-4}$  

Đáp án:

a) $S=\left\{ \dfrac{10}{3} \right\}$

b) $S=\left\{ \dfrac{1}{4};-2 \right\}$

c) $S=\left\{ 5 \right\}$

d) $S=\left\{ 7 \right\}$

Giải thích các bước giải:

a) $8x-3=5x+7$

$\Leftrightarrow 8x-5x=7+3$

$\Leftrightarrow 3x=10$

$\Leftrightarrow x=\dfrac{10}{3}$

Vậy $S=\left\{ \dfrac{10}{3} \right\}$

b) $\left( 4x-1 \right)\left( x+2 \right)=0$

$\Leftrightarrow 4x-1=0$ hoặc $x+2=0$

$\Leftrightarrow 4x=1$ hoặc $x=-2$

$\Leftrightarrow x=\dfrac{1}{4}$ hoặc $x=-2$

Vậy $S=\left\{ \dfrac{1}{4};-2 \right\}$

c) $\dfrac{2x-1}{3}+\dfrac{x+4}{2}=\dfrac{5x+20}{6}$

$\Leftrightarrow \dfrac{2\left( 2x-1 \right)}{6}+\dfrac{3\left( x+4 \right)}{6}=\dfrac{5x+20}{6}$

$\Leftrightarrow 4x-2+3x+12=5x+20$

$\Leftrightarrow 4x+3x-5x=20+2-12$

$\Leftrightarrow 2x=10$

$\Leftrightarrow x=5$

Vậy $S=\left\{ 5 \right\}$

d) $\dfrac{3}{x-2}+\dfrac{1}{x+2}=\dfrac{5x-3}{{{x}^{2}}-4}\,\,\,\,\,\left( * \right)$

ĐKXĐ: $x\ne \pm 2$

MTC: $\left( x-2 \right)\left( x+2 \right)$

$\left( * \right)\Leftrightarrow \dfrac{3\left( x+2 \right)}{\left( x-2 \right)\left( x+2 \right)}+\dfrac{x-2}{\left( x-2 \right)\left( x+2 \right)}=\dfrac{5x-3}{\left( x-2 \right)\left( x+2 \right)}$

$\Leftrightarrow 3x+6+x-2=5x-3$

$\Leftrightarrow 3x+x-5x=-3-6+2$

$\Leftrightarrow -x=-7$

$\Leftrightarrow x=7$ (nhận)

Vậy $S=\left\{ 7 \right\}$