Lời giải:

a) ĐKXĐ: \(x\geq 8\)

\(A=\sqrt{x+1}-\sqrt{x-8}=\frac{9}{\sqrt{x+1}+\sqrt{x-8}}\). Mà:

\(\sqrt{x+1}+\sqrt{x-8}=\sqrt{(\sqrt{x+1}+\sqrt{x-8})^2}=\sqrt{2x-7+2\sqrt{(x+1)(x-8)}}\)

\(\geq \sqrt{2.8-7+2.0}=3\) với mọi $x\geq 8$

Do đó: \(A=\frac{9}{\sqrt{x+1}+\sqrt{x-8}}\leq \frac{9}{3}=3\)

Vậy \(A_{\max}=3\Leftrightarrow x=8\)

b) ĐKXĐ: \(3\leq x\leq 5\)

\(B=\sqrt{x-3}+\sqrt{5-x}=\sqrt{(\sqrt{x-3}+\sqrt{5-x})^2}=\sqrt{2+2\sqrt{(x-3)(5-x)}}\)

\(\geq \sqrt{2+2.0}=\sqrt{2}, \forall 3\leq x\leq 5\)

Vậy \(B_{\min}=\sqrt{2}\Leftrightarrow 3\leq x\leq 5\)

\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)

\(A_{max}=10\) khi \(\dfrac{\sqrt{x-1}}{3}=\dfrac{\sqrt{5-x}}{4}\Rightarrow x=\dfrac{61}{25}\)

\(A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\left(\sqrt{x-1}+\sqrt{5-x}\right)\ge3\sqrt{x-1+5-x}=6\)

\(A_{min}=6\) khi \(x=5\)

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 \((1994-x)(x+1995)\geq 0\Leftrightarrow -1995\leq x\leq 1994\)

Bài 3:

Ta thấy:

\(2x-x^2+7=8-(x^2-2x+1)=8-(x-1)^2\leq 8, \forall x\in\mathbb{R}\)

\(\Rightarrow 2+\sqrt{2x-x^2+7}\leq 2+\sqrt{8}=2+2\sqrt{2}\)

\(\Rightarrow B=\frac{3}{2+\sqrt{2x-x^2+7}}\geq \frac{3}{2+2\sqrt{2}}\)

Vậy GTNN của $B$ là \(\frac{3}{2+2\sqrt{2}}\).

Đẳng thức xảy ra tại \((x-1)^2=0\Leftrightarrow x=1\)

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 \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

1:

a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

Lời giải:

Đặt \(\sqrt{x}=a(a\ge 0)\)

Khi đó: \(P=\frac{4a}{3(a^2-a+1)}\)

Để \(P=\frac{8}{9}\Rightarrow \frac{4a}{3(a^2-a+1)}=\frac{8}{9}\)

\(\Rightarrow \frac{a}{a^2-a+1}=\frac{2}{3}\Rightarrow 3a=2(a^2-a+1)\)

\(\Leftrightarrow 2a^2-5a+2=0\Leftrightarrow (a-2)(2a-1)=0\)

\(\Rightarrow \left[\begin{matrix} a-2=0\\ 2a-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} a=2=\sqrt{x}\\ a=\frac{1}{2}=\sqrt{x}\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x=4\\ x=\frac{1}{4}\end{matrix}\right.\) (t/m)

b)

Vì \(a\geq 0; a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}>0\)

Do đó: \(P=\frac{4}{3}.\frac{a}{a^2-a+1}\geq \frac{4}{3}.0=0\)

Vậy \(P_{\min}=0\Leftrightarrow a=0\Leftrightarrow x=0\)

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Áp dụng BĐT Cô-si: \(a^2+1\geq 2a\Rightarrow a^2-a+1\geq 2a-a=a\)

\(\Rightarrow \frac{a}{a^2-a+1}\leq \frac{a}{a}=1\Rightarrow P=\frac{4}{3}.\frac{a}{a^2-a+1}\leq \frac{4}{3}.1=\frac{4}{3}\)

Vậy \(P_{\max}=\frac{4}{3}\Leftrightarrow a=1\Leftrightarrow x=1\)

Bài 2:

Đặt \(P=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-\sqrt{12-4\sqrt{5}}\)

Có:

\(4+\sqrt{15}=\frac{8+2\sqrt{15}}{2}=\frac{5+3+2\sqrt{3.5}}{2}=\frac{(\sqrt{3}+\sqrt{5})^2}{2}\)

\(\Rightarrow \sqrt{4+\sqrt{15}}=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{2}}\)

Tương tự: \(\sqrt{4-\sqrt{15}}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

\(12-4\sqrt{5}=12-2\sqrt{20}=10+2-2\sqrt{10.2}=(\sqrt{10}-\sqrt{2})^2\)

\(\Rightarrow \sqrt{12-4\sqrt{5}}=\sqrt{10}-\sqrt{2}\)

Vậy \(P=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{2}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-(\sqrt{10}-\sqrt{2})\)

\(=\sqrt{2}\)

\(A^2=\left(2\sqrt{x-4}+\sqrt{8-x}\right)^2\le\left(2^2+1^2\right)\left(x-4+8-x\right)=20..\)

\(A\le2\sqrt{5}..\)

Bài a, c tìm GTLN thì làm được rồi, chỉ không biết tìm GTNN bằng BĐT như thế nào?