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

\(\Leftrightarrow A=|x|-\sqrt{\left(x-2\right)^2}\)

\(\Leftrightarrow A=x-|x-2|=x-x+2=2\)

A = \(\sqrt{x^2}-\sqrt{x^2-4x+4}=\sqrt{x^2}-\sqrt{\left(x-2\right)^2}=\left|x\right|-\left|x-2\right|=x-x+2=2\)(vì  \(x\ge2\))

B = \(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=\left|x-3\right|-\left|x+3\right|=3-x+x+3=6\)(vì x < 3)

Áp dụng BĐT côsi ta có:

\(VT\le\frac{a^2+1-b^2}{2}+\frac{b^2+1-c^2}{2}+\frac{c^2+1-a^2}{2}=\frac{3}{2}\)

Đẳng thức đề bài chỉ xảy ra khi \(a=b=c=\frac{\sqrt{2}}{2}\)

=> \(a^2+b^2+c^2=\frac{3}{2}\)(ĐPCM)

Ta có \(x-1=\sqrt[3]{2}+\sqrt[3]{4}\)

<=> \(\left(x-1\right)^3=6+3.\sqrt[3]{2.4}.\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)

<=>\(x^3-3x^2+3x-1=6+6.\left(x-1\right)\)

<=>\(x^3-3x^2-3x-1=0\)

=> \(P=x^2\left(x^3-3x^2-3x-1\right)-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x-1+2016\)

=> \(P=2016\)

grtfgvgg

\(A=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)

\(A=\sqrt{9+6\sqrt{5}+5}+\sqrt{9-6\sqrt{5}+5}\)

 \(A=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(A=3+\sqrt{5}+3-\sqrt{5}=6\)

b) \(B=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(B=\sqrt{3-4\sqrt{3}+4}-\sqrt{3+4\sqrt{3}+4}\)

\(B=\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)

\(B=2-\sqrt{3}-\sqrt{3}-2=-2\sqrt{3}\)

Câu a tách 14 thành 5+9 . Có hằng đẳng thức

Câu b tương tự tách 7 thành 4+ 3 nhé

Bài làm:

a) \(\left(2x-1\right)x^2\ge0\), mà \(x^2\ge0\)

\(\Rightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)

b) \(3+2x>0\Leftrightarrow2x>-3\Leftrightarrow x>-\frac{3}{2}\)

c) \(4-5x\ge0\Leftrightarrow4\ge5x\Rightarrow x\le\frac{4}{5}\)

d) \(\left(x-3\right)\left(x+3\right)\ge0\)nên ta xét 2 TH sau:

+ Nếu: \(\hept{\begin{cases}x-3\ge0\\x+3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\ge-3\end{cases}}\Rightarrow x\ge3\)

+ Nếu: \(\hept{\begin{cases}x-3\le0\\x+3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\le-3\end{cases}}\Rightarrow x\le-3\)

Vậy \(\orbr{\begin{cases}x\ge3\\x\le-3\end{cases}}\)

\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

Bài 2:

 a, Ta có 

   \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}\)

= \(3\left|-2\right|+\left|-5\right|\)

=\(6+5\)

= 11

Vậy \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}=11\)

b, Ta có 

     \(\sqrt{6+2\sqrt{5}}-\sqrt{5}\)

=  \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5}\)

=   \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}\)

=    \(\left|\sqrt{5}+1\right|-\sqrt{5}\)

=    \(\sqrt{5}+1-\sqrt{5}=1\)

Vậy \(\sqrt{6+2\sqrt{5}}-\sqrt{5}=1\)