a) \(A=4x\left(x+y\right)-5y\left(x-y\right)-4x^2=4x^2+4xy-5xy+5y^2-4x^2=5y^2-xy\)

Với x = -5; y = 2 thì: \(A=5\cdot2^2-\left(-5\right)\cdot2=20+10=30\)

b) \(B=-3x\left(x^2+y^2\right)+2y\left(x^2-y\right)=-3x^3-3xy^2+2yx^2-2y^2=-3x^3+2x^2y-3xy^2-2y^2\)

Với x = 1; y = 2 thì: \(B=-3\cdot1^3+2\cdot1^2\cdot2-3\cdot1\cdot2^2-2\cdot2^2=-3+4-12-8=-19\)

Viết các đơn thức sau dưới dạng thu gọn 

a)3x.5y².x²

\(A=4x\left(x+y\right)-5y\left(x-y\right)-4x^2\)

     \(=4x^2+4xy-5y^2-5xy-4x^2\)

      = \(\left(4x^2-4x^2\right)+\left(4xy-5xy\right)-5y^2\)

       \(=5y^2-xy\)

Thay x=-5 và y=2 vào đa thức \(5y^2-xy\) ta được:

\(5.2^2-\left(-5\right).2=20+10=30\)

Vậy 30 là giá trị của đa thức trên tại x=-5 và y=2

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

    \(=-3x^3-3xy^2+2yx^2-2y^2\)

    \(=-3x^3-3xy^2+2yx^2-2y^2\)

Thay x=1 và y=2 vào đa thức \(=-3x^3-3xy^2+2yx^2-2y^2\)

\(\left(-3\right).1^3-2.1.2^2+2.2.1^2-2.2^2=-3-8+4-8=-15\)

Vậy -15 là giá trị của đa thức \(=-3x^3-3xy^2+2yx^2-2y^2\)  tại x=1 và y=2

^...^ ^_^ hihihi

\(a,\left(x+2\right)^2-\left(x-2\right)^2-2\left(x-2\right)\left(x+2\right).\)

\(=\left(x+2-x+2\right)^2=4^2=16\)

\(b,\left(x-y\right)^2+\left(x+y\right)^2+2\left(x-y\right)\left(x+y\right)\)

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

\(c,\left(x-y+z\right)^2-2\left(x+y\right)-2\left(x+y\right)\left(x-y\right)-z^2\)

a, (\(x\) + y).(\(x\) + y)² - 3\(xy\).(\(x\) + y) 

= (\(x+y\))³ - 3\(x^2\)y - 3\(xy^2\)

= \(x^3\) + 3\(x^2\).y + 3\(xy^2\) + y³ - 3\(x^2\).y  - 3\(xy^2\)

= \(x^3\) + y³ 

b, (\(x-y\)).(\(x-y\))² - 3\(xy\).(\(x-y\)) 

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

= \(x^3\) - 3\(x^2\)y + 3\(xy^2\) - y³ - 3\(x^2\)y + 3\(xy^2\)

= \(x^3\) - 6\(x^2\)y + 6\(xy^2\) - y³

c, (\(x\) - 2y)² + 4y²

=  \(x^2\) - 4\(xy\) + 4y² + 4y²

=  \(x^2\) - 4\(xy\) + 8y²

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

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

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

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

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

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

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

A=\(\frac{-y\left(1-y\right)+x\left(1-y\right)\left(1+y\right)}{\left(1-y\right)}\)

A=\(\frac{\left(1-y\right)\left(-y+x+xy\right)}{1-y}\)=\(x-y+xy\)

B1

a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)

b, \(=>B=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left[x+y-x+y\right]^2=\left[2y\right]^2=4y^2\)

c,\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

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

d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a-b+c\right)^2-\left(b-c\right)^2\)

\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)

\(+\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)\)

\(=a\left(a+2b-2c\right)+a\left(a-2b\right)\)

\(=a\left(a+2b-2c+a-2b\right)=a\left(2a-2c\right)=2a^2-2ac\)

B2:

\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)

\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)

\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)

Bài 1: 

a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

=4xy

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

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

\(=\left(2y\right)^2=4y^2\)

c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

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

\(=\left(x^3-1\right)\left(x^3+1\right)\)

\(=x^6-1\)

d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)

\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a+b+c\right)^2-\left(b-c\right)^2\)

\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)

\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)

\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)

\(=2a^2-4bc\)

Bài 2: 

Ta có: x+y=3

nên \(\left(x+y\right)^2=9\)

\(\Leftrightarrow2xy+17=9\)

\(\Leftrightarrow2xy=-8\)

hay xy=-4

Ta có: \(x^3+y^3\)

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

\(=3^3-3\cdot\left(-4\right)\cdot3\)

\(=27+36=63\)

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

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

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

\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)

\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)

\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)

\(=-5x-27\)

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

\(=8x^3+y^3-\left(8x^3-y^3\right)\)

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

\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x+y+z-x-y\right)^2\)

\(=z^2\)

a)

=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\) 

=-5x-19

b)

=\(8x^3+y^3-8x^3+y^3\) 

=\(2y^3\) 

c)

=(x+y+z-x-y)\(^2\) +x+y

=\(z^2+x+y\) 

hc tốt