Bài 5: Phép cộng các phân thức đại số

Bài 8 mình lm rồi nhé

Bài 9:

a) \(=\dfrac{3x-2-7x+4}{2xy}=\dfrac{-4x+2}{2xy}=\dfrac{2\left(-2x+1\right)}{2xy}=-\dfrac{2x-1}{xy}\)

b) \(=\dfrac{3x+2-\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)

a) \(=\dfrac{3x+1+x^2-6x}{x^2-3x+1}=\dfrac{x^2-3x+1}{x^2-3x+1}=1\)

b) \(=\dfrac{6x+5x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{6x^2+18x}{\left(x-3\right)\left(x+3\right)}=\dfrac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{6x}{x-3}\)

\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-8}{x^2-4}=\dfrac{2\left(x^2-4\right)}{x^2-4}=2\)

\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-2\right)+x-14}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x+2x+2+x^2-2x-2x+4+x-14}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{2x^2-8}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{2\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =2\)

\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)

bài 5 thôi nhé mn

6)

\(\dfrac{3}{2x}+\dfrac{3x-3}{2x-1}+\dfrac{2x^2+1}{4x^2-2x}\)

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

\(=\dfrac{3\left(2x-1\right)+\left(3x-3\right).2x+2x^2+1}{2x\left(2x-1\right)}\)

\(=\dfrac{6x-3+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}\)

\(=\dfrac{8x^2-2}{2x\left(2x-1\right)}\)

\(=\dfrac{2\left(4x^2-1\right)}{2x\left(2x-1\right)}\)

\(=\dfrac{2\left(2x+1\right)\left(2x-1\right)}{2\left(2x-1\right)}\)

\(=2x+1\)

\(=\dfrac{2a}{\left(a-5\right)^2}+\dfrac{10}{\left(a-5\right)\left(a+5\right)}=\dfrac{2a+10+10a-50}{\left(a-5\right)^2\left(a+5\right)}=\dfrac{12a-40}{\left(a-5\right)^2\left(a+5\right)}\)

A nên đăng vào mấy h mà nhìu ng onl í 

chứ h này ko còn ai đou ạ

\(=\dfrac{x+2+x+x-2}{x\left(x+2\right)}=\dfrac{3x}{x\left(x+2\right)}=\dfrac{3}{x+2}\)

\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)