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\(A=1+\sqrt{x-2}\)
Do \(\sqrt{x-2}\ge0\forall x>2\) nên \(A\ge1\forall x>2\)
Vậy \(minA=1\Leftrightarrow x=2\)
__________
\(B=5-\sqrt{2x-1}\)
Do \(\sqrt{2x-1}\ge0\forall x\ge\frac{1}{2}\)nên \(B\le5\forall x\ge\frac{1}{2}\)
Vậy \(maxB=5\Leftrightarrow x=\frac{1}{2}\)
a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)
\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)
b. \(0\le\sqrt{4-x^2}\le2\)
\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)
vậy \(GTNN=\frac{\sqrt{46}}{4}\)
ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)
\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)
Lời giải:
Áp dụng BĐT AM-GM:
$y\sqrt{x-1}=\sqrt{y^2(x-1)}=\sqrt{y(xy-y)}\leq \frac{y+xy-y}{2}=\frac{xy}{2}$
$x\sqrt{y-2}=\sqrt{x^2(y-2)}=\sqrt{x(xy-2x)}\leq \frac{2x+(xy-2x)}{2\sqrt{2}}=\frac{xy}{2\sqrt{2}}$
$\Rightarrow y\sqrt{x-1}+x\sqrt{y-2}\leq \frac{xy}{2}+\frac{xy}{2\sqrt{2}}=xy.\frac{2+\sqrt{2}}{4}$
$\Rightarrow P\leq \frac{2+\sqrt{2}}{4}$
Vậy $P_{\max}=\frac{2+\sqrt{2}}{4}$
Tìm GTLN của biểu thức:
a. \(A=\dfrac{1}{x-\sqrt{x}+1}\)
b. \(B=\dfrac{2x-2\sqrt{x}+5}{x-\sqrt{x}+2}\)
\(\text{Condition}:1-2x-2x^2\ge0\)
We have:
\(A=x+\sqrt{1-2x-2x^2}\)
\(\Rightarrow M=-2A=-2x-2\sqrt{1-2x-2x^2}\)
Now we need to find min of M
We have it:
\(M=-2x-2\sqrt{1-2x-2x^2}=\left(1-2x-2x^2-2\sqrt{1-2x-2x^2}+1\right)+2x^2-2=\left(\sqrt{1-2x-2x^2}-1\right)^2+2x^2-2\ge-2\)
\(\Rightarrow-2A\ge-2\Leftrightarrow A\le1\)
Sign '=' happening when \(x=0\)
\(B=\sqrt{-x^2+2x+2}=\sqrt{-x^2+2x-1+3}\)
\(=\sqrt{-\left(x^2-2x+1\right)+3}\)
\(=\sqrt{-\left(x-1\right)^2+3}\le\sqrt{3}\)
Xảy ra khi x=1
Ta có \(-x^2+2x+2\)=\(-\left(x^2-2x+1-3\right)\)=\(3-\left(x-1\right)^2\le3\)
Dấu '=' xảy ra \(\Leftrightarrow x=1\)
Do đó MaxB=\(\sqrt{3}\)(Dấu '=' xảy ra \(\Leftrightarrow x=1\))
`B=1-\sqrt{x^2-2x+2}`
`=1-sqrt{x^2-2x+1+1}`
`=1-sqrt{(x-1)^2+1}`
Vì `(x-1)^2>=0=>(x-1)^2+1>=1`
`=>sqrt{(x-1)^2+1}>=1`
`=>B<=1-1=0`
Dấu "=" xảy ra khi `x-1=0<=>x=1`
Vậy `GTLN_B=0<=>x=1.`