giải PT sau
a) 2x-4=0
b) (x-5)(x+6)=0
c) 2x-3=2(2x-3)
d) $\frac{x-2}{x+3}$ = $\frac{x+1}{x-2}$
giải PT sau a) 2x-4=0 b) (x-5)(x+6)=0 c) 2x-3=2(2x-3) d) $\frac{x-2}{x+3}$ = $\frac{x+1}{x-2}$
By Eva
By Eva
giải PT sau
a) 2x-4=0
b) (x-5)(x+6)=0
c) 2x-3=2(2x-3)
d) $\frac{x-2}{x+3}$ = $\frac{x+1}{x-2}$
a,
$ 2x – 4 = 0 \\ ⇔ 2x = 4 \\ \rightarrow x = 2 $
Vậy $ S = { 2 } $
b,
$ (x-5)(x+6)=0 $ $\\$ $ ⇔ x – 5 = 0 $ hoặc $x + 6 = 0 $ $\\$ $ \rightarrow x=5$ hoặc $x = -6$
Vậy $ S = { 5 ; -6 }$
c,
$2x-3=2(2x-3)$ $\\$ $ ⇔2x – 3 = 4x – 6$ $\\$ $ ⇔ -2x = -3$ $\\$ $ \rightarrow x= \dfrac{3}{2}$
Vậy $S = { \dfrac{3}{2} }$
d,
$\dfrac{x-2}{x+3} = \dfrac{x+1}{x-2}$ $\\$ $ ⇔ (x-2)(x-2) = (x+1)(x+3)$ $\\$ $ ⇔ x^2 – 2x – 2x + 4 = x^2 + 3x + x + 3 $ $\\$ $ ⇔ -8x = -1 \\ \rightarrow x = \dfrac{1}{8}$
Vậy $ S = { \dfrac{1}{8} } $