tính
a. \(\frac{x+1}{2x+6}\)+ \(\frac{2x+3}{x\left(x+3\right)}\)
b. \(\frac{x-1}{x}\)+ \(\frac{x+2}{2}\)
c. \(\frac{1}{x+y}\)+ \(\frac{-1}{x-y}\)+ \(\frac{2x}{x^2+y^2}\)
d. \(\frac{1}{\left(x-y\right)\left(y-z\right)}\)+ \(\frac{1}{\left(z-x\right)\left(x-y\right)}\)+ \(\frac{1}{\left(y-z\right)\left(z-x\right)}\)với x, y, z từng đôi một khác nhau
e. \(\frac{3x^2-6xy}{x^3-y^3}\)+ \(\frac{3y^2}{x^3-y^3}\)
giải giúp mình với
a) \(\frac{x+1}{2x+6}\)+\(\frac{2x+3}{x\left(x+3\right)}\)
= \(\frac{x+1}{2\left(x+3\right)}\)+ \(\frac{2x+3}{x\left(x+3\right)}\)
= \(\frac{x\left(x+1\right)}{2x\left(x+3\right)}\)+ \(\frac{2\left(2x+3\right)}{2x\left(x+3\right)}\)
= \(\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
= \(\frac{x^2+5x+6}{2x\left(x+3\right)}\)
= \(\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
= \(\frac{x+2}{2x}\)
b) \(\frac{x-1}{x}\)+ \(\frac{x+2}{2}\)
= \(\frac{2\left(x-1\right)}{2x}\)+ \(\frac{x\left(x+2\right)}{2x}\)
= \(\frac{2x-2+x^2+2x}{2x}\)
= \(\frac{x^2+4x-2}{2x}\)
c) \(\frac{1}{x+y}\)+ \(\frac{-1}{x-y}\)+ \(\frac{2x}{x^2+y^2}\)
= \(\frac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+\(\frac{-\left(x+y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+ \(\frac{2x\left(x-y\right)\left(x+y\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)
= \(\frac{x^3+xy^2-x^2y-y^3-x^3-xy^2-xy^2-y^3+2x^3+2x^2y-2x^2y+2xy^2}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2x^3+xy^2-x^2y-2y^3}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{\left(2x^3-2y^3\right)-\left(x^2y-xy^2\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2\left(x-y\right)\left(x^2+xy+y^2\right)-xy\left(x-y\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{\left(x-y\right)\left(2x^2+2xy+2y^2-xy\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2x^2+xy+2y^2}{\left(x+y\right)\left(x^2+y^2\right)}\)
e) = \(\frac{3x^2-6xy+3y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
= \(\frac{3\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
=\(\frac{3x-3y}{x^2+xy+y^2}\)
( Mình bận rồi, lát làm câu d nhé)