Question

[ミ★Ƥɦươŋɠ Ňɦї★彡]

TÍNH:

a)   \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)

b)   \(\frac{3-2x}{x^2-9}+\frac{1}{x^2-6}\)

c)  \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)

d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)

Answer 1

[Edogawa Conan]

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}=\frac{6.2}{2x\left(x+4\right)}+\frac{3x}{2x\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(x+4\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)

c) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5.y}{2y\left(y+2\right)}+\frac{2\left(y-2\right)}{2y\left(y+2\right)}=\frac{-5y+2y-4}{2y\left(y+2\right)}=\frac{-3y-4}{2y\left(y+2\right)}\)

d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x\left(x-2y\right)}-\frac{3}{x\left(x-2y\right)}=\frac{x-1-3}{x\left(x-2y\right)}=\frac{x-4}{x\left(x-2y\right)}\)