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\(R=x\sqrt{3-x^2}=\sqrt{x^2\left(3-x^2\right)}\le\dfrac{x^2+3-x^2}{2}=\dfrac{3}{2}\)
\(\Rightarrow R_{max}=\dfrac{3}{2}\) khi \(x^2=3-x^2\Rightarrow x=\dfrac{\sqrt{6}}{2}\)
\(R=x\sqrt{3-x^2}\le\frac{x^2+3-x^2}{2}=\frac{3}{2}\)
đạt được khi \(x=\sqrt{\frac{3}{2}}\)
Bài 2 :
b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)
ĐKXĐ : \(x\ge1\)
Pt(1) tương đương :
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)
Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)
\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\) ( Thỏa mãn )
Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)
Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\Leftrightarrow2=2\) ( Luôn đúng )
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)
Bài 1 :
a) ĐKXĐ : \(-1\le a\le1\)
Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)
\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)
\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)
\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)
Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)
b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :
\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)
Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
\(\Rightarrow0\le x< \frac{9}{4}\)
c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)
Vậy \(MinR=-3\Leftrightarrow x=0\)
Ta có:
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2}{2}\)
\(P=\left(-\sqrt{x}\right)\left(\sqrt{x}-1\right)\)
\(P=\sqrt{x}-x\)
b) Để \(P>0\) thì \(\sqrt{x}-x>0\)
- \(\sqrt{x}-x>0\)
\(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
Suy ra: TH1: \(\sqrt{x}< 0\) và \(1-\sqrt{x}< 0\) (Loại) vì \(\sqrt{x}\ge0\)
TH2:\(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) (Nhận)
Ta có \(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) để \(P>0\)
- \(\sqrt{x}>0\) \(\Rightarrow x>0\)
- \(1-\sqrt{x}>0\) \(\Rightarrow\sqrt{x}< 1\) \(\Rightarrow x< 1\)
Vậy để \(P>0\) thì \(0< x< 1\)
c)\(P=\sqrt{x}-x\)
\(P=-\left(x-\sqrt{x}\right)\)
\(P=-\left(\left(\sqrt{x}\right)^2-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)
\(P=-\left(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\right)\)
\(P=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)
Nên \(-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) \(\Rightarrow x=\frac{1}{4}\)
Vậy GTLN của \(P\) là \(\frac{1}{4}\) khi \(x=\frac{1}{4}\)
R=\(x\sqrt{3-x^2}=\sqrt{x^2\left(3-x^2\right)}\)
AD BĐT cô si \(a+b\ge2\sqrt{ab}\)
=>\(R=\sqrt{x^2\left(3-x^2\right)}\le\dfrac{x^2+3-x^2}{2}=\dfrac{3}{2}\)
Vậy GTLN của R=\(\dfrac{3}{2}\Leftrightarrow x^2=3-x^2\Leftrightarrow x=\dfrac{\sqrt{6}}{2}\)