\(=2y\cdot\dfrac{5y}{4}\cdot\dfrac{16}{25x^2y^3}=\dfrac{8}{5x^2y}\)

\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

3x(2x-y)-3x²

=6x²-3xy-3x²

=3x²-3xy

\((x+4)(x-2)-(x-3)^2\)

\(=(x⋅x+x⋅(−2)+4⋅x+4⋅(−2))−(x⋅x+x⋅(−3)+(−3)⋅x+(−3)⋅(−3))\)

\(=(x^2-2x+4x-8)-(x^2-3x-3x+9)\)

\(\) \(=(x^2+2x-8)-(x^2-6x+9)\)

\(=x^2+2x-8-x^2+6x-9\)

\(=8x−17\)

(x+4)(x-2)-(x-3)^2

=x^2-2x+4x-8-x^2+6x-9

=2x-8-9

=2x-17

tichs ddi bn owi

bcbcvxvbx

Ta có:

 VT=(x2+y2)2−(2xy)2VT=(x2+y2)2−(2xy)2

=(x2+y2−2xy)(x2+y2+2xy)=(x2+y2−2xy)(x2+y2+2xy)

=(x−y)2(x+y)2=VP=(x−y)2(x+y)2=VP

⇒đpcm⇒đpcm

Bexiu bị j ấy

`@` `\text {Ans}`

`\downarrow`

`(x^2 + 2)^2`

`= x^4 + 4x^2 + 4`

____

`@` Áp dụng: `(A+B)^2 = A^2 + 2AB + B^2.`

\(\left(x^2+2\right)^2\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2\)

\(=x^4+4x^2+4\)

\(\left(8x+7y\right)\left(2x^2-xy+9\right)\)

\(=8x\left(2x^2-xy+9\right)+7y\left(2x^2-xy+9\right)\)

\(=16x^3-8x^2y+72x+14x^2y-7xy^2+63y\)

\(=16x^3+6x^2y-xy^2+72x+63y\)

thanhs