Với mọi số thực dương x;y;z ta có:

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2xy+2yz+2zx\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Leftrightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)

Áp dụng:

a.

\(\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\le\sqrt{3\left(a+2+b+2+c+2\right)}=\sqrt{3\left(21+6\right)}=9\)

b.

\(\sqrt{a+b+2}+\sqrt{b+c+2}+\sqrt{c+a+2}\le\sqrt{3\left(a+b+2+b+c+2+c+a+2\right)}\)

\(\Rightarrow\sqrt{a+b+2}+\sqrt{b+c+2}+\sqrt{c+a+2}\le\sqrt{6\left(a+b+c\right)+18}=\sqrt{6.21+18}=12\)

Dấu "=" xảy ra khi \(a=b=c=7\)

\(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\)

\(\Leftrightarrow\sqrt{7-2\sqrt{7}\sqrt{3}+3}=\sqrt{a}-\sqrt{b}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{a}-\sqrt{b}\)

\(\Leftrightarrow\sqrt{7}-\sqrt{3}=\sqrt{a}-\sqrt{b}\)

=>a=7;b=3 =>a-b=7-3=4 

ko bik đúng ko

bài này a,b phải nguyên thì may ra lm đc

\(\sqrt{10-2\sqrt{21}}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)}\)

=> a=7; b=3

a-b=4

\(\sqrt{10-2\sqrt{21}}=\sqrt{3-2\sqrt{3}.\sqrt{7}+7}=\sqrt{7}-\sqrt{3}\Rightarrow a-b=4\)

Hoàng Anh Tú câu này dễ òm mà , giải mò cái j

\(\sqrt{10-2\sqrt{21}}\)= \(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)=/ \(\sqrt{7}\)+ \(\sqrt{3}\)/ (giá trị tuyệt đối /)= \(\sqrt{7}\)+ \(\sqrt{3}\) ( do \(\sqrt{7}\)+\(\sqrt{3}\) >0)

=> \(\sqrt{a}\)+ \(\sqrt{b}\)= \(\sqrt{7}\)+ \(\sqrt{3}\)

=> a+b= 7+3=10

câu a) \(\sqrt{5+2\sqrt{6}}+\sqrt{14-4\sqrt{6}}\)

GG