`8.`
`a)x²-6x+10`
`=x²-6x+9+1`
`=(x²-6x+9)+1`
`=(x²-2.x.3+3²)+1`
`=(x-3)²+1`
Ta có:`(x-3)²≥0∀x`
`⇒(x-3)²+1≥1>0∀x`
`⇒x²-6x+10>0∀x(đpcm)`
`b)4x-x²-5`
`=-(x²-4x+5)`
`=-(x²-4x+4+1)`
`=-(x²-4x+4)-1`
`=-(x²-2.x.2+2²)-1`
`=-(x-2)²-1`
Ta có:`(x-2)²≥0∀x`
`⇒-(x-2)²≤0∀x`
`⇒-(x-2)²-1≤-1<0∀x`
`⇒4x-x²-5<0∀x(đpcm)`
`9.`
`a)P=x²-2x+5`
`=x²-2x+1+4`
`=(x²-2x+1)+4`
`=(x-1)²+4`
Ta có:`(x-1)²≥0∀x`
`⇒(x-1)²+4≥4∀x`
Vậy `P_(min)=4` khi `x-1=0⇔x=1`
`b)Q=2x²-6x`
`=2(x²-3x)`
`=2(x²-3x+9/4-9/4)`
`=2(x²-3x+9/4)-9/2`
`=2[x²-2.x. 3/2+(3/2)^2]-9/2`
`=2(x-3/2)^2-9/2`
Ta có:`(x-3/2)^2≥0∀x`
`⇒2(x-3/2)^2≥0∀x`
`⇒2(x-3/2)^2-9/2≥-9/2∀x`
Vậy `Q_(min)=-9/2` khi `x-3/2=0⇔x=3/2`
`c)M=x²+y²-x+6y+10`
`=x²+y²-x+6y+1/4+9+3/4`
`=(x²-x+1/4)+(y²+6y+9)+3/4`
`=[x²-2.x. 1/2+(1/2)^2]+(y²+2.y.3+3²)+3/4`
`=(x-1/2)^2+(y+3)²+3/4`
Ta có:`(x-1/2)^2≥0∀x`
`(y+3)^2≥0∀y`
`⇒(x-1/2)^2+(y+3)^2≥0∀x,y`
`⇒(x-1/2)^2+(y+3)²+3/4≥3/4∀x,y`
Dấu `'='` xảy ra khi$\begin{cases} x-\dfrac{1}{2}=0\\y+3=0 \end{cases}$`⇔`$\begin{cases} x=\dfrac{1}{2}\\y=-3 \end{cases}$
Vậy `M_(min)=3/4` khi `x=1/2` và `y=-3`
