Question

Answer 1

gấp

gấp

gấp

Answer 2

\(\dfrac{x}{x+1}-\dfrac{3}{x-2}-\dfrac{x^2+x}{x^2-x-2}=0\)

\(\Leftrightarrow\dfrac{x}{x+1}-\dfrac{3}{x-2}-\dfrac{x^2+x}{\left(x+1\right)\left(x-2\right)}=0\left(dkxd:x\ne-1,x\ne2\right)\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x+1\right)-x^2-x=0\)

\(\Leftrightarrow x^2-2x-3x-3-x^2-x=0\)

\(\Leftrightarrow-6x=3\)

\(\Leftrightarrow x=-2\left(tmdk\right)\)

Vậy \(S=\left\{-2\right\}\)

 

Answer 3

ĐKXĐ: x ≠ -1; x ≠ 2

Phương trình đã cho tương đương:

x(x - 2) - 3(x + 1) - x² - x = 0

⇔ x² - 2x - 3x - 3 - x² - x = 0

⇔ -6x - 3 = 0

⇔ -6x = 3

⇔ x = -1/2 (nhận)

Vậy S = {-1/2}