Áp dụng bđt bu - nhi -a, ta có 

\(A^2\le\left(3^2+1\right)\left(x^2+2-x\right)=20\Rightarrow A\le2\sqrt{5}\)

dấu = xayra <=>\(\frac{x}{3}=\sqrt{2-x^2}\Leftrightarrow9\left(2-x^2\right)=x^2\Leftrightarrow18=10x^2\Leftrightarrow x=\frac{3}{\sqrt{5}}\)

Bạn ơi, mình chưa hiểu chỗ này lắm, tại sao \(\left(3^2\text{+}1\right)\left(x^2\text{+}2-x\right)\text{=}20\)

Xem câu hỏi

Ta có: \(A=\sqrt{x-1-2\sqrt{x-2}}-\sqrt{x+7-6\sqrt{x-2}}\)

\(=\sqrt{x-2-2\cdot\sqrt{x-2}\cdot1+1}-\sqrt{x-2-6\cdot\sqrt{x-2}+9}\)

\(=\sqrt{\left(\sqrt{x-2}-1\right)^2}-\sqrt{\left(\sqrt{x-2}-3\right)^2}=\left|\sqrt{x-2}-1\right|-\left|\sqrt{x-2}-3\right|\)

=>\(A\le\left|\sqrt{x-2}-1-\sqrt{x-2}+3\right|=2\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x-2}-1\right)\left(\sqrt{x-2}-3\right)\ge0\)

=>\(\left[\begin{array}{l}\sqrt{x-2}\ge3\\ \sqrt{x-2}\le1\end{array}\right.\Rightarrow\left[\begin{array}{l}x-2\ge9\\ 0\le x-2\le1\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge11\\ 2\le x\le3\end{array}\right.\)

a: Đặt \(x^2-4=a\)

Pt sẽ là \(a=3\sqrt{xa}\)

\(\Rightarrow a^2=9xa\)

\(\Leftrightarrow a\left(a-9x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)

hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)

d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)

Pt sẽ là 2a+b=ab+2

=>(b-2)(1-a)=0

=>b=2 và 1-a

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

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

GTLN = 3

ÔI, em nhầm rùi

= -(\(\sqrt{x}\)+ 1/2) +1/4 +2 

GTLN = 9/4

Lần này không sai đâu chị

ĐKXĐ: \(\left\{{}\begin{matrix}x-2>=0\\4-x>=0\\x+1< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2< =x< =4\\x< >-1\end{matrix}\right.\Leftrightarrow x\in\left[2;4\right]\)

\(x=1\)

1 ) \(A=\sqrt{x-2}+\sqrt{4-x}\)

ĐKXĐ : \(2\le x\le4\)

\(\Rightarrow A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)

Áp dụng bđt AM - GM ta có : 

\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\)

\(\Rightarrow A^2\le2+2=4\Rightarrow-2\le A\le2\)

Mà A > 0 nên ko thể có min = - 2 nên \(2\le x\le4\) ta chọn x = 2

=> A = \(\sqrt{2}\)

Vậy \(\sqrt{2}\le A\le2\)