Question

Giải hộ mình vs ạ !

Answer 1

a) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\) (1)

ĐKXĐ: x≠ 2

(1) ⇔ \(\dfrac{2x-10}{6\left(x-2\right)}-\dfrac{3x-6}{6\left(x-2\right)}=\dfrac{6x-9}{6\left(x-2\right)}\)

⇒ 2x - 10 - 3x + 6 = 6x - 9

⇔ -7x = -5

⇔ x = \(\dfrac{5}{7}\) (TMĐKXĐ)

Vậy S=\(\left\{\dfrac{5}{7}\right\}\)

b)\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\) (2)

ĐKXĐ: x≠ \(\pm2\)

(2) ⇒ x² - 3x +2 - x² - 2x = 5x - 2

⇔ -10x = -4

⇔ x = \(\dfrac{2}{5}\) (TMĐKXĐ)

Vậy S= \(\left\{\dfrac{2}{5}\right\}\)

c) \(\dfrac{6}{x+5}-\dfrac{1}{5-x}=\dfrac{3x+5}{x^2-25}\) (3)

ĐKXĐ: x ≠ \(\pm5\)

(3) ⇒ 6x - 30 -x +5 = 3x + 5

⇔ 2x = 30

⇔ x = 15 (TMĐKXĐ)

Vậy S= \(\left\{15\right\}\)

Answer 2

d) ĐKXĐ: x≠ \(\pm2\)

⇒ x² - 3x + 2 - x² - 2x = 2 - 5x

⇔ 0x = 0 (TMĐKXĐ)

Vậy PT có vô số nghiệm

e) ĐKXĐ: x≠ 0; x≠ 2

⇒ x² + 2x = x - 2 + 2 

⇔ x² + x = 0

⇔ x(x + 1) = 0

⇔\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy S = \(\left\{-1\right\}\)

f) ĐKXĐ: x≠ \(\pm2\)

⇒ x² - 2x + x + 2 = 2 - 3x

⇔ x² + 4x = 0

⇔ x(x+4) = 0

⇔\(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-4\left(TMĐKXĐ\right)\end{matrix}\right.\)\(\)

Vậy S=\(\left\{-4;0\right\}\)