`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`

\(-1\le sin\left(x^2\right)\le1\Rightarrow\)\(0\le\sqrt{1-sin\left(x^2\right)}\le\sqrt{2}\Rightarrow-1\le y\le\sqrt{2}-1\)

\(y_{min}=-1\) khi \(sin\left(x^2\right)=1\Rightarrow x=\pm\sqrt{\dfrac{\pi}{2}+k2\pi}\) (\(k\in N\))

\(y_{max}=\sqrt{2}-1\) khi \(sin\left(x^2\right)=-1\Rightarrow x=\pm\sqrt{-\dfrac{\pi}{2}+k2\pi}\) (\(k\in Z^+\))

Làm như thế nào ạ

\(L=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)

=>\(L>=2\sqrt{\left(\sqrt{x}-1\right)\cdot\dfrac{1}{\sqrt{x}-1}}+2=4\)

Dấu = xảy ra khi (căn x-1)^2=1

=>căn x-1=1 hoặc căn x-1=-1

=>x=0 hoặc x=4

\(P=\dfrac{x+3}{\sqrt{x}+3}\) (ĐK: \(x\ge0\))

Mà: \(x\ge0\Rightarrow\left\{{}\begin{matrix}x+3\ge3\\\sqrt{x}+3\ge3\end{matrix}\right.\) nên:

\(P=\dfrac{x+3}{\sqrt{x}+3}\ge\dfrac{3}{3}=1\)

Dấu "=" xảy ra:

\(\dfrac{x+3}{\sqrt{x}+3}=1\)

\(\Leftrightarrow x=\sqrt{x}\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy: \(P_{min}=1\) khi \(x=0\)