Question

[Phan thanh hằng]

A, 4/x+2 + 3/x-2 + 5x+2/4-x^2

B, 1/x-y + 3xy/y^3-x^3 + x-y/x^2+xy+y^2

Answer 1

[Edogawa Conan]

a) Ta có: \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}\)

\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2}{x+2}\)

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

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

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

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

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

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