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Câu 1:
Tìm max:
Áp dụng BĐT Bunhiacopxky ta có:
\(y^2=(3\sqrt{x-1}+4\sqrt{5-x})^2\leq (3^2+4^2)(x-1+5-x)\)
\(\Rightarrow y^2\leq 100\Rightarrow y\leq 10\)
Vậy \(y_{\max}=10\)
Dấu đẳng thức xảy ra khi \(\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}\Leftrightarrow x=\frac{61}{25}\)
Tìm min:
Ta có bổ đề sau: Với $a,b\geq 0$ thì \(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
Chứng minh:
\(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
\(\Leftrightarrow (\sqrt{a}+\sqrt{b})^2\geq a+b\)
\(\Leftrightarrow \sqrt{ab}\geq 0\) (luôn đúng).
Dấu "=" xảy ra khi $ab=0$
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Áp dụng bổ đề trên vào bài toán ta có:
\(\sqrt{x-1}+\sqrt{5-x}\geq \sqrt{(x-1)+(5-x)}=2\)
\(\sqrt{5-x}\geq 0\)
\(\Rightarrow y=3(\sqrt{x-1}+\sqrt{5-x})+\sqrt{5-x}\geq 3.2+0=6\)
Vậy $y_{\min}=6$
Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-1)(5-x)=0\\ 5-x=0\end{matrix}\right.\Leftrightarrow x=5\)
Bài 2:
\(A=\sqrt{(x-1994)^2}+\sqrt{(x+1995)^2}=|x-1994|+|x+1995|\)
Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:
\(A=|x-1994|+|x+1995|=|1994-x|+|x+1995|\geq |1994-x+x+1995|=3989\)
Vậy \(A_{\min}=3989\)
Đẳng thức xảy ra khi \((1994-x)(x+1995)\geq 0\Leftrightarrow -1995\leq x\leq 1994\)
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)
\(A=\frac{\sqrt{x}-2}{x-4}\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Lời giải:
Áp dụng BĐT Cauchy cho các số dương ta có:
\(\sqrt{y-2}\leq \frac{(y-2)+2}{2\sqrt{2}}=\frac{y}{2\sqrt{2}}\) \(\Rightarrow x\sqrt{y-2}\leq \frac{xy}{2\sqrt{2}}\)
\(\sqrt{x-3}\leq \frac{(x-3)+3}{2\sqrt{3}}=\frac{x}{2\sqrt{3}}\Rightarrow y\sqrt{x-3}\leq \frac{xy}{2\sqrt{3}}\)
Do đó:
\(M=\frac{x\sqrt{y-2}+y\sqrt{x-3}}{xy}\leq \frac{\frac{xy}{2\sqrt{2}}+\frac{xy}{2\sqrt{3}}}{xy}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\)
Vậy \(M_{\max}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\) khi \(x=6;y=4\)
\(\sqrt{y-2}\le\frac{y-2+1}{2}chứ?\)
cách cô em không hiểu ??
ĐKXĐ : \(\hept{\begin{cases}x^2-2\ge0\\7-x^2\ge0\end{cases}}\Leftrightarrow\sqrt{2}\le x\le\sqrt{7}\)
Áp dụng bất đẳng thức Bunhiacopxki
Ta có N = \(\sqrt{x^2+1}+\sqrt{2\left(x^2-2\right)}+\sqrt{3\left(7-x^2\right)}\)
\(=1.\sqrt{x^2+1}+1.\sqrt{2\left(x^2-2\right)}+1.\sqrt{3\left(7-x^2\right)}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left[x^2+1+2\left(x^2-2\right)+3\left(7-x^2\right)\right]}\)
\(=\sqrt{3.18}=\sqrt{54}\)
Dấu "=" xảy ra <=> \(\frac{1}{x^2+1}=\frac{1}{2\left(x^2-2\right)}=\frac{1}{3\left(7-x^2\right)}\)
<=> x2 + 1 = 2x2 - 4
<=> x = \(\sqrt{5}\)(tm)
Vậy Max N = \(\sqrt{54}\Leftrightarrow x=\sqrt{5}\)