Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(1^2+2^2\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)

\(\Rightarrow5\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)

\(\Rightarrow5\left(x^2+4y^2\right)\ge1^2=1\)

\(\Rightarrow5\left(x^2+4y^2\right)\ge\dfrac{1}{5}\)

Đẳng thức xảy ra khi \(x=y=\dfrac{1}{5}\)

a) `x^2+y^2-2x+4y+5`

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

`=(x-1)^2+(y+2)^2 >=0 forall x,y`

b) `-3x^2+2x-5`

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

`=-[(\sqrt3 x)^2 -2.\sqrt3 x .\sqrt3/3 + (\sqrt3/3)^2 +14/5]`

`=-(\sqrt3 x-\sqrt3/3)^2-14/5 < 0 forall x`

b) Ta có: \(-3x^2+2x-5\)

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

\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{14}{9}\right)\)

\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{3}< 0\forall x\)

1) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x=25\Leftrightarrow x=\dfrac{25}{2}\left(tm\right)\)

2) \(=\sqrt{\dfrac{\dfrac{1}{4}}{9}}=\dfrac{\dfrac{1}{2}}{3}=\dfrac{1}{6}\)

3) \(=\sqrt{225a^2}=15a\left(do.a\ge0\right)\)

4) \(=2y^2.\dfrac{x^2}{2\left|y\right|}=\left[{}\begin{matrix}x^2y\left(y>0\right)\\-x^2y\left(y< 0\right)\end{matrix}\right.\)