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

\(=x^5y^4-2x^3y^7-2x^7y^3+2x^3y^7\)

\(=x^5y^4-2x^7y^3\)

`5x^2 (x-y) - 15xy(y-x) `

`= 5x^2 (x-y) + 15xy(x-y) `

`= (5x^2 + 15xy)(x-y) `

`= 5x(x + 3y)(x-y) `

`(x+y)^2 - 6(x+y) + 9` (sửa đề)

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

`x^2 - 5x + 6`

`= x^2 - 2x - 3x +6`

`= (x^2 - 2x) - (3x - 6) `

`= x(x-2) - 3(x-2) `

`= (x-3)(x-2) `

Ta có:

`x^2+4x+1`

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

`=(x+2)^2-3`

`(x+2)^2>=0` với mọi x

`=>(x+2)^2-3>=-3` với mọi x 

Dấu "=" xảy ra: 

`x+2=0<=>x=2` 

Vậy: ...

Ta có:

\(x^2+4x+1\\ =x^2+4x+2-1\\ =\left(x+2\right)^2-1\)

Vì: \(\left(x+2\right)^2\ge0\rightarrow\left(x+2\right)^2-1\ge-1\forall x\)

Vậy: GTNN là: \(-1\) 

\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=-7\\ \Rightarrow x^2-4x+4-\left(x^2-2x-3\right)=-7\\ \Rightarrow x^2-4x+4-x^2+2x+3=-7\\ \Rightarrow-2x+7=-7\\ \Rightarrow-2x=-14\\ \Rightarrow x=-14:\left(-2\right)\\ \Rightarrow x=7\)

Cậu ơi thuộc CD vs cái nào nữa k hay M vs N thuộc mỗi CD thôi ?

M với N thuộc với mỗi CD thui nhé ạ

\(\left(5-x\right)^2+\left(x+5\right)^2-2\cdot\left(x+5\right)\cdot\left(x-5\right)\\=\left(x-5\right)^2-2\cdot\left(x+5\right)\cdot\left(x-5\right)+\left(x+5\right)^2\\ =\left(x-5-x-5\right)^2\\ =\left(-10\right)^2\\ =100\)

\(a,\left(9x^2y^3+6x^3y^2-4xy^2\right):3xy^2\\ =9x^2y^3:3xy^2+6x^3y^2:3xy^2-4xy^2:3xy^2\\ =3xy+2x^2-\dfrac{4}{3}\\ b,\dfrac{1}{2}xy\left(x^5-y^3\right)-x^2y\left(\dfrac{1}{4}x^4-y^3\right)\\ =\dfrac{1}{2}xy\cdot x^5-\dfrac{1}{2}xy\cdot y^3-x^2y\cdot\dfrac{1}{4}x^4+x^2y\cdot y^3\\ =\dfrac{1}{2}x^6y-\dfrac{1}{2}xy^4-\dfrac{1}{2}xy^4-\dfrac{1}{4}x^6y+x^2y^4\\ =\dfrac{1}{4}x^6y-\dfrac{1}{2}xy^4+x^2y^4\)

Với mọi x;y;z ta có:

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-2\right)^2\ge0\\\left(z-2\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)

\(\Rightarrow x^2+y^2+z^2\ge4\left(x+y+z\right)-12\) (1)

Đồng thời cũng có:

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\) 

\(\Rightarrow4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+zx\right)\)(2)

Cộng vế (1) và (2):

\(\Rightarrow5\left(x^2+y^2+z^2\right)\ge4\left(x+y+z+xy+yz+zx\right)-12=4.18-12=60\)

\(\Rightarrow x^2+y^2+z^2\ge\dfrac{60}{5}=12\)

Dấu "=" xảy ra khi \(x=y=z=2\)

a) Ta có:

`(n-3)(n+3)-(n-1)(n+9)`

`=(n^2-3^2)-(n^2-n+9n-9)`

`=(n^2-9)-(n^2+8n-9)`

`=n^2-9-n^2-8n+9`

`=-8n` chia hết cho 8 

b) Ta có:

`(n+7)(n+5)-(n+1)(n-1)`

`=(n^2+7n+5n+35)-(n^2-1^2)`

`=n^2+12n+35-n^2+1`

`=12n+36`

`=12(n+3)` chia hết cho 12