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a: ĐKXĐ: \(x\ge\dfrac{1}{3}\)
b: ĐKXĐ: \(x< \dfrac{15}{2}\)
c: ĐKXĐ: \(x\le0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ Tìm Max. Ta có
\(\frac{M}{2}=\frac{15x}{2}+\frac{x\sqrt{17-x^2}}{2}\)
\(=-\left(\frac{x^2}{16}-\frac{2x\sqrt{17-x^2}}{4}+17-x^2\right)-15\left(\frac{x^2}{16}-\frac{2x}{4}+1\right)+32\)
\(=-\left(\frac{x}{4}-\sqrt{17-x^2}\right)^2-15\left(\frac{x}{4}-1\right)^2+32\le32\)
\(\Rightarrow M\le64\)
\(\Rightarrow\)GTLN là M = 64 đạt được khi x = 4
Tìm Min. Ta có
\(\frac{M}{2}=\frac{15x}{2}+\frac{x\sqrt{17-x^2}}{2}\)
\(=\left(\frac{x^2}{16}+\frac{2x\sqrt{17-x^2}}{4}+17-x^2\right)+15\left(\frac{x}{16}+\frac{2x}{4}+1\right)-32\)
\(=\left(\frac{x}{4}+\sqrt{17-x^2}\right)^2+15\left(\frac{x}{4}+1\right)^2-32\ge-32\)
\(\Rightarrow M\ge-64\)
Vậy GTNN là M = - 64 đạt được khi x = - 4
![](https://rs.olm.vn/images/avt/0.png?1311)
Với các số thực không âm a; b ta luôn có BĐT sau:
\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) (bình phương 2 vế được \(2\sqrt{ab}\ge0\) luôn đúng)
Áp dụng:
a.
\(A\ge\sqrt{x-4+5-x}=1\)
\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(A\le\sqrt{\left(1+1\right)\left(x-4+5-x\right)}=\sqrt{2}\) (Bunhiacopxki)
\(A_{max}=\sqrt{2}\) khi \(x-4=5-x\Leftrightarrow x=\dfrac{9}{2}\)
b.
\(B\ge\sqrt{3-2x+3x+4}=\sqrt{x+7}=\sqrt{\dfrac{1}{3}\left(3x+4\right)+\dfrac{17}{3}}\ge\sqrt{\dfrac{17}{3}}=\dfrac{\sqrt{51}}{3}\)
\(B_{min}=\dfrac{\sqrt{51}}{3}\) khi \(x=-\dfrac{4}{3}\)
\(B=\sqrt{3-2x}+\sqrt{\dfrac{3}{2}}.\sqrt{2x+\dfrac{8}{3}}\le\sqrt{\left(1+\dfrac{3}{2}\right)\left(3-2x+2x+\dfrac{8}{3}\right)}=\dfrac{\sqrt{510}}{6}\)
\(B_{max}=\dfrac{\sqrt{510}}{6}\) khi \(x=\dfrac{11}{30}\)
a)Ta có:A=\(\sqrt{x-4}+\sqrt{5-x}\)
=>A2=\(x-4+2\sqrt{\left(x-4\right)\left(5-x\right)}+5-x\)
=>A2= 1+\(2\sqrt{\left(x-4\right)\left(5-x\right)}\ge1\)
=>A\(\ge\)1
Dấu '=' xảy ra <=> x=4 hoặc x=5
Vậy,Min A=1 <=>x=4 hoặc x=5
Còn câu b tương tự nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\sqrt{2\left(x+1\right)}}{\sqrt{2}}+\frac{\sqrt{2\left(y+1\right)}}{\sqrt{2}}+\frac{\sqrt{2\left(z+1\right)}}{\sqrt{2}}\le\frac{x+y+z+9}{2\sqrt{2}}=3\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge0\)
\(Q=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}=3+\dfrac{6}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{6}{\sqrt{x}+3}\le2\Rightarrow3+\dfrac{6}{\sqrt{x}+3}\le5\)
\(\Rightarrow Q_{max}=5\) khi \(x=0\)
\(Q-5=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}-5=\dfrac{-2\sqrt{x}}{\sqrt{x}+3}\le0\)
\(\Rightarrow Q\le5\)
\(Q_{maxx}=5\) khi \(\sqrt{x}=0\Leftrightarrow x=0\)