(2x-5 +5+2x)(2x-5-5-2x) = 0

-40x =0

x=0

x²-3x+5x-15-x²-2x 

=-15

  \(\left(x-3\right).\left(x+5\right)-x.\left(x+2\right)\)

\(=x^2+5x-3x-15-x^2-2x\)

\(=-15\)

\(\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)

bài như này hả bạn

`(3(x-y)^5-2(x-y)^4+3(x-y)^2)/(5(x-y)^2)`

`=((x-y)^2(3(x-y)^3-2(x-y)^2+3)/(5(x-y)^2)`

`=(3(x-y)^3-2(x-y)^2+3)/5`

`=3/5(x-y)^3-2/5(x-y)^2+3/5.`

A= -x² + 5

x² ≥ 0 ⇔ -x² ≤ 0 ⇔ A + 5 ≤ 5 ⇔ A(max ) = 5 ⇔ x =0

Đặt `A=-x^2+5`

Ta có: `x^2>=0`

`->-x^2<=0`

`->\underbrace{-x^2+5}_{A}<=5`

Dấu "=" xảy ra khi `x=0`

Vậy `A_max=5<=>x=0.`

\(x^2-x+2022\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{8087}{4}\)

\(=\left(x-\dfrac{1}{4}\right)^2+\dfrac{8087}{4}\ge\dfrac{8087}{4}\)

`(2x-1)(4x^2+2x+1)`

`=(2x-1)[(2x)^2+2x.1+1^2]`

`=(2x)^3-1^3`

`=8x^3-1.`

= (2x)³ - 1^{3 =}  8x³ - 1

   (X² - 3)(X⁴ +3X²+9) 

=  (X² -3){ (X²)² + 3.X² + 3²)}

= X³ - 27

      2022² - 88 x 1011 + 484

=  2² x 1011² - 44 x 1011 -  44 x 1011 + 484

= 1011 x (4 x 1011 - 44) - 44 x (1011 -11)

= 1011 x 4000 -  44 x 1000

= 1011 x 4 x 1000 - 44 x 1000

= 1000 x ( 1011 x 4 - 44)

= 1000 x 4000

= 4000000

Lời giải:
$x^2-4x+8=(x^2-4x+4)+4=(x-2)^2+4$

Ta thấy: $(x-2)^2\geq 0$ với mọi $x$ nên $x^2-4x+8=(x-2)^2+4\geq 4>0$ với mọi $x$ 

Ta có đpcm.