a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

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

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

c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)

=1/3

d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)

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

a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6

a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

kha sdaif dòng mik xin phép trình bày bằng lời ạ :

a) tìm MTC rồi quy đồng lên làm bình thường ại , tử cộng tử mấu giữ nguyên 

b) cx vậy ạ tách mẫu tìm MTC rồi ....

~ hok tốt ~

a)\(\dfrac{2x^2-10xy}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)

\(=\dfrac{2x\left(x-5y\right)}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)

\(=\dfrac{x-5y}{y}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)

\(=\dfrac{x\left(x-5y\right)+x\left(5y-x\right)+y\left(x+2y\right)}{xy}\)

\(=\dfrac{x^2-5xy+5xy-x^2+xy+2y^2}{xy}\)

\(=\dfrac{y\left(x+2y\right)}{xy}\)

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

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

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

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

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

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

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

e) \(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)

\(=\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\)

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

\(=\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\)

\(=\dfrac{x^2-2xy+y^2}{x-y}\)

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

\(=x-y\)

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

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

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

d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)

\(=x^2+x+1-x+1=x^2+2\)

Sao ảnh đại diện của bạn giống mình thế?