Question

chứng tỏ:

a)x^2+y^2-2x+4y+5 lớn hơn hoặc bằng 0

b)-3x^2+2x-5 nhỏ hơn 0

Answer 1

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`

Answer 2

b) Ta có: \(-3x^2+2x-5\)

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

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

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