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.`

\(\dfrac{3\left(x-y\right)^{5}-2\left(x-y\right)^{4}+3\left(x-y\right)^{2}}{5\left(x-y\right)^{2}} \left(1\right)\\\)

Đặt `x-y=a`

\(\left(1\right)\Leftrightarrow\dfrac{3a^{5}-2a^{4}+3a^{2}}{5a^{2}}\\=\dfrac{a^{2}\left(3a^{3}-2a^{2}+3\right)}{5a^{2}}\\=\dfrac{3}{5}a^{3}-\dfrac{2}{5}a^{2}+\dfrac{3}{5} \ \left(2\right)\)Thay `x-y=a` 

vào \(\left(2\right)\)

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

\(\dfrac{3\left(x-y\right)^{5}-2\left(x-y\right)^{4}+3\left(x-y\right)^{2}}{5\left(x-y\right)^{2}} \left(1\right)\)

Đặt `x-y=a`

\(\left(1\right)<=>\dfrac{3a^{5}-2a^{4}+3a^{2}}{5a^{2})\\=\dfrac{a^{2}\left(3a^{3}-2a^{2}+3\right)}{5a^{2}}\\=dfrac{3}{5}a^{3}-\dfrac{2}{5}a^{2}+\dfrac{3}{5} \left(2\right)\)

Thay `x-y=a` vào \(\left(2\right)\)

\(\left(2\right)<=>dfrac{3}{5}\left(x-y\right)^{3}-\dfrac{2}{5}\left(x-y\right)^{2}+\dfrac{3}{5}.\)

`(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.`

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

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

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

`=8x^3-1.`