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`A=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`
`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`
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a) \(N=13-6x-3x^2\)
\(-N=3x^2+6x-13\)
\(-N=3\left(x^2+2x+1\right)-16\)
\(-N=3\left(x+1\right)^2-16\)
Mà \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow-N\ge-16\)
\(\Leftrightarrow N\le16\)
Dấu "=" xảy ra khi : \(x+1=0\Leftrightarrow x=-1\)
Vậy ...
b) \(Q=2x^2-8x+11\)
\(Q=2\left(x^2-4x+4\right)+3\)
\(Q=2\left(x-2\right)^2+3\)
Mà \(\left(x-2\right)^2\ge0\forall x\) \(\Rightarrow2\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow Q\ge3\)
Dấu "=" xảy ra khi : \(x-2=0\Leftrightarrow x=2\)
Vậy ...
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a: ĐKXĐ: x>=0; x<>1
b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)
d: căn x+2>=2
=>A<=1/2
Dấu = xảy ra khi x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
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