K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

23 tháng 6 2021

\(\dfrac{x^2-9y^2}{x^2+xy-6y^2}=\dfrac{\left(x-3y\right)\left(x+3y\right)}{\left(x-2y\right)\left(x+3y\right)}=\dfrac{x-3y}{x-2y}\)

5 tháng 12 2023

\(A=(xy^2-1)(x^2y+5)-xy^2(x^2y+5)\\=(x^2y+5)(xy^2-1-xy^2)\\=(x^2y+5)\cdot(-1)\\=-x^2y-5\)

A=( xy^2-1)(x^2y+5)-xy^2( x^2y+5)

=xy^2.(x^2y+5)-1.(x^2y+5)-x^3y^3-5xy^2

=-x^2-5 ( bước này làm tắc )

 

a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

\(=3x-y-y-x+2x^2-2x\)

\(=2x^2-2y\)

27 tháng 12 2020

\(\dfrac{x^2-xy-6y^2}{x^2-9y^2}=\dfrac{\left(x-3y\right)\left(x+2y\right)}{\left(x-3y\right)\left(x+3y\right)}=\dfrac{x+2y}{x+3y}\)

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

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

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

\(=\dfrac{1}{x-y}\)

29 tháng 7 2023

a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)

\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)

\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)

\(Q=\left(x-y-2x-4y\right)^2\)

\(Q=\left(-x-5y\right)^2\)

b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)

\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)

\(A=\left[\left(xy+2\right)-2\right]^3\)

\(A=\left(xy+2-2\right)^3\)

\(A=\left(xy\right)^3\)

\(A=x^3y^3\)

c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)

\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)

\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)

\(=0\)

a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2

=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2

b: =(xy+2-2)^3=(xy)^3=x^3y^3

c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)

=24x+2x^3-2x^3-24x

=0

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

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

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

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