Question

[TSO_Đức Đạt]

Tính a) 2x/x+5 + 10/x+5 b) x+2/x-2 - x-2/x+2 + 16/x^2-4

Answer 1

[Phạm Hồ Thanh Quang]

a)\(\frac{2x}{x+5}+\frac{10}{x+5}=\frac{2x+10}{x+5}=\frac{2\left(x+5\right)}{x+5}=2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{8\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{x-2}\)

Answer 2

[Phong Du]

a) \(\frac{2x}{x+5}+\frac{10}{x+5}\)=\(\frac{2x+10}{x+5}\)=\(\frac{2\left(x+5\right)}{x+5}\)=\(2\)

b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}\)=\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2-x+2\right)\left(x+2+x-2\right)+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4\times2x+16}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8}{x-2}\)