Thiếu câu C :(

a: \(\left(2x^2y-3xy+4xy^2\right):2xy\)

\(=\frac{2x^2y}{2xy}-\frac{3xy}{2xy}+\frac{4xy^2}{2xy}\)

\(=x-\frac32+2y\)

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

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

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

c: \(\left\lbrack\frac{x}{xy-y^2}-\frac{2x-y}{x^2-xy}\right\rbrack:\left(\frac{1}{x}-\frac{1}{y}\right)\)

\(=\left\lbrack\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x\left(x-y\right)}\right\rbrack:\frac{y-x}{xy}\)

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

=-1

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

a) \(x^2y\left(5xy-2x^2y-y^2\right)\)

\(=5x^3y^2-2x^4y^2-x^2y^3\)

b) \(\left(x-2y\right)\left(2x^3+4xy\right)\)

\(=2x^4+4x^2y-4x^3y-8xy^2\)

a) (𝑥−5)(𝑥+7)

𝑥(𝑥+7)−5(𝑥+7)

𝑥²+7𝑥−5(𝑥+7)

𝑥²+7𝑥−5𝑥−35

𝑥²+2𝑥−35

b) (𝑥−3)(𝑥²+5𝑦)

𝑥(𝑥²+5𝑦)−3(𝑥²+5𝑦)

𝑥³+5𝑥𝑦−3(𝑥²+5𝑦)

𝑥³+5𝑥𝑦−3𝑥²−15𝑦

𝑥³−3𝑥²+5𝑥𝑦−15𝑦

`a)[3x+2]/[x^2]:[6x+4]/[2x^2]`       

`=[3x+2]/[x^2].[2x^2]/[2(3x+2)]`

`=1`

____________________________________________________

`b)[4xy]/[x+y]:[6x^2y^3]/[x^2-y]`         

`=[4xy]/[x+y].[(x-y)(x+y)]/[6xy.xy^2]`

`=[2(x-y)]/[3xy^2]=[2x-2y]/[3xy^2]`

\(x^4.\frac{2x^2-x-1}{3}\)

đúng k bn

đúng rồi bạn Thanh Ngân ơi

Giups mình nhé

\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)

\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)

all ạ

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

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

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