Đặt \(P=\sqrt{21-2x}+\sqrt{2x-3}\)

\(\Rightarrow P^2=\left(1.\sqrt{21-2x}+1.\sqrt{2x-3}\right)^2\)

\(\le\left(1^2+1^2\right)\left[\left(\sqrt{21-2x}\right)^2+\left(\sqrt{2x-3}\right)^2\right]\)

\(=2.18=36\)

\(\Rightarrow P\le6\)

Dấu "=" xảy ra khi \(21-2x=2x-3\Leftrightarrow x=6\)

Vậy GTLN của biểu thức đã cho là 6.

Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\sqrt{21-2x}+\sqrt{2x-3})^2\leq (21-2x+2x-3)(1+1)=36$

$\Rightarrow \sqrt{21-2x}+\sqrt{2x-3}\leq 6$

Vậy GTLN của biểu thức là $6$. Giá trị này đạt được khi:

$21-2x=2x-3\Leftrightarrow x=6$

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

đk x² - 2x \(\ge\) 0  => x \(\in\) (-\(\infty\); 0] \(\cup\) [ 2; + \(\infty\))

\(\sqrt{x^2-2x}\) \(\ge\) 0

- \(\sqrt{x^2-2x}\) \(\le\) 0 

A \(\le\) 3 => A(max) = 3 <=> x² - 2x = 0 \(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(y=2cos^2x-2\sqrt{3}sinx.cosx+1\)

\(=2cos^2x-1-2\sqrt{3}sinx.cosx+2\)

\(=cos2x-\sqrt{3}sin2x+2\)

\(=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)

\(=2cos\left(2x+\dfrac{\pi}{3}\right)+2\)

Ta có: \(cos\left(2x+\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)

\(\Rightarrow min=0\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=-1\Leftrightarrow2x+\dfrac{\pi}{3}=\pi+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

\(\Rightarrow max=4\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=1\Leftrightarrow2x+\dfrac{\pi}{3}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

\(y=2cos^2x-\sqrt{3}sin2x+1=cos2x-\sqrt{3}sin2x+2\)

\(y=2.cos\left(2x+\dfrac{\pi}{3}\right)+2\)

\(\forall x\in R->-1\le cos\left(2x+\dfrac{\pi}{3}\right)\)

=> \(Min_y=2.\left(-1\right)+2=0\) 

Mặt khác, theo Bunhiacopxki:

\(\left(cos2x+\sqrt{3}sin2x\right)^2\le\left(1^2+\sqrt{3}^2\right)\left(cos^22x+sin^22x\right)=4\)

=>\(Max_y=4\)

a) Để A có nghĩa :

\(\Rightarrow\sqrt{2x+3-x^2\: }\Leftrightarrow2+\sqrt{2x+3-x^2}\ge2\forall x\) 

\(\Rightarrow\sqrt{-\left(x-1\right)^2+4}\ge0\) 

\(\Leftrightarrow-\left(x-1\right)^2\ge-4\) 

\(\Leftrightarrow\left(x-1\right)^2\le4\) 

\(\Rightarrow3\ge x\ge-1\) 

Vậy.....

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)