`I=x²+5y²-4xy+2y+6`
`=x²+y²+4y²-4xy+2y+1+5`
`=(x^2-4xy+4y^2)+(y^2+2y+1)+5`
`=[x^2-2.x.2y+(2y)^2]+(y^2+2.y.1+1^2)+5`
`=(x-2y)^2+(y+1)^2+5`
Ta có:`(x-2y)²≥0∀x,y`
`(y+1)²≥0∀y`
`⇒(x-2y)^2+(y+1)^2≥0∀x,y`
`⇒(x-2y)^2+(y+1)^2+5≥5∀x,y`
Dấu `'='` xảy ra khi$\begin{cases} x-2y=0\\y+1=0 \end{cases}$`⇔`$\begin{cases} x=-2\\y=-1 \end{cases}$
Vậy `I_(min)=5` khi `x=-2` và `y=-1`
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`K=4x²-4x+9`
`=4x²-4x+1+8`
`=(4x²-4x+1)+8`
`=[(2x)²-2.2x.1+1²]+8`
`=(2x-1)²+8`
Ta có:`(2x-1)²≥0∀x`
`⇒(2x-1)²+8≥8∀x`
Vậy `K_(min)=8` khi `2x-1=0⇔2x=1⇔x=1/2`
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`L=25x²-5x-1`
`=25x²-5x+1/4-5/4`
`=(25x²-5x+1/4)-5/4`
`=[(5x)²-2.5x. 1/2+(1/2)^2]-5/4`
`=(5x-1/2)^2-5/4`
Ta có:`(5x-1/2)^2≥0∀x`
`⇒(5x-1/2)^2-5/4≥-5/4∀x`
Vậy `L_(min)=-5/4` khi `5x-1/2=0⇔5x=1/2⇔x=1/10`
