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

=[(x+y)+(x-y)]²

=[(x+y)+(x-y)]²

^=(2x)2

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

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

\(=6x^2y+2y^3\)

Cách khác:

Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

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

\(=6x^2y+2y^3\)

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

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

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

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

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

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

a,hđt số 3 = \(\left(a^2+2a\right)^2-9\) 

b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)

a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)

\(=\left(a^2+2a\right)^2-9\)

b) \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)

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

a, 0

b,0

c, 0

mình ko chắc lắm

a/ (x+y)(x+y)

   =x+y.x+y

   =x+x.y+y

   =2.x.2.y

    =2.(x+y)

\(=x^3+3x^2.5y+3x.25y^2+125y^3-\left(8x^6-3.4x^4+3.2x^2y^2-y^3\right)\)

\(=2x^3+15x^2y+75xy^2+125y^3-8x^6+12x^4-6x^2y^2\)

Mình lm luôn k ghi đề nhé

\(\left(x+5y\right)^3-\left(2x^2-y\right)^3\)

\(=x^3+15x^2y+75xy^2+125y^3-8x^6+12x^4y-6x^2y^2+y^3\)

\(=8x^6+x^3+15x^2y+75xy^2+126y^3+12x^4y-6x^2y^2\)

Linh Linh bạn làm sai kìa

`@` `\text {Ans}`

`\downarrow`

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

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

`= x^2 - xy + xy - y^2 + y^2 - x^2`

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

`= 0`