Đáp án:
`a)A_{min}=4` khi `x=1`
`b)B_{min}=3/4` khi `x=1/2`
`c)C_{min}=-36` khi `x=0;x=-5`
`d)D_{min}=2` khi `x=y=-1/2`
Giải thích các bước giải:
`a)A=x^2-2x+5`
`=x^2-2x+1+4`
`=(x^2-2x+1)+4`
`=(x^2-2x.1+1^2)+4`
`=(x-1)^2+4`
Ta có:`(x-1)^2≥0∀x`
`⇒(x-1)^2+4≥4∀x`
Vậy `A_{min}=4` khi `x-1=0⇔x=1`
`b)B=x^2-x+1`
`=x^2-2.x. 1/2+1/4+3/4`
`=(x^2-2.x. 1/2+1/4)+3/4`
`=[x^2-2.x. 1/2+(1/2)^2]+3/4`
`=(x-1/2)^2+3/4`
Ta có:`(x-1/2)^2≥0∀x`
`⇒(x-1/2)^2+3/4≥3/4∀x`
Vậy `B_{min}=3/4` khi `x-1/2=0⇔x=1/2`
`c)C=(x-1)(x+2)(x+3)(x+6)`
`=[(x-1)(x+6)][(x+2)(x+3)]`
`=(x^2+6x-x-6)(x^2+3x+2x+6)`
`=(x^2+5x-6)(x^2+5x+6)`
`=(x^2+5x)^2-6^2`
`=(x^2+5x)^2-36`
Ta có:`(x^2+5x)^2≥0∀x`
`⇒(x^2+5x)^2-36≥-36∀x`
Vậy `C_{min}=-36` khi `x^2+5x=0⇔x(x+5)=0⇔`$\left[\begin{matrix} x=0\\ x+5=0\end{matrix}\right.$`⇔`\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
`d)D=x^2+5y^2-2xy+4y+3`
`=x^2-2xy+y^2+4y^2+4y+1+2`
`=(x^2-2xy+y^2)+(4y^2+4y+1)+2`
`=(x-y)^2+[(2y)^2+2.2y.1+1^2]+2`
`=(x-y)^2+(2y+1)^2+2`
Ta có:`(x-y)^2≥0∀x,y`
`(2y+1)^2≥0∀y`
`⇒(x-y)^2+(2y+1)^2≥0∀x,y`
`⇒(x-y)^2+(2y+1)^2+2≥2∀x,y`
Dấu `'='` xảy ra khi$\begin{cases} x-y=0\\\\2y+1=0 \end{cases}$`⇔`$\begin{cases} x=y\\\\y=\dfrac{-1}{2} \end{cases}$`⇔`$\begin{cases} x=\dfrac{-1}{2}\\\\y=\dfrac{-1}{2} \end{cases}$
Vậy `D_{min}=2` khi `x=y=-1/2`
