\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

\(A^2=4+2\sqrt{4-3}=4+2=6\)

\(A=\sqrt{6}\)

\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(\sqrt{2}A=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

\(\sqrt{2}A=\sqrt{\left(1+\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)

\(\sqrt{2}A=1+\sqrt{3}+\sqrt{3}-1=2\sqrt{3}\)

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

Xem câu hỏi

\(P=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+1}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{xz}+\sqrt{z}+1}\)( Vì xyz=1 nên \(\sqrt{xyz}=1\))

\(P=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{y}+1+\sqrt{yz}\right)}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{z}\left(\sqrt{x}+1+\sqrt{xy}\right)}\)

\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{1}{\sqrt{x}+1+\sqrt{xy}}\)

\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{\sqrt{xyz}}{\sqrt{x}\left(1+\sqrt{yz}+\sqrt{y}\right)}\)

\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{\sqrt{yz}}{\sqrt{y}+1+\sqrt{yz}}=\frac{\sqrt{y}+1+\sqrt{yz}}{\sqrt{y}+1+\sqrt{yz}}=1\)

help me

Sai đề rồi

ko sai nhé

Áp dụng BĐT Cauchy-Schwarz dạng ENgel ta có:

\(VT=\frac{3}{ab+bc+ca}+\frac{2}{a^2+b^2+c^2}\)

\(=\frac{\sqrt{6}^2}{2\left(ab+bc+ca\right)}+\frac{\sqrt{2}^2}{a^2+b^2+c^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)

\(=\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(a+b+c\right)^2}\approx15>14\)

có em!

\(\sqrt{7-4\sqrt{3}}\)=\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{\left(2-\sqrt{3}\right)^2}\)= \(\left|2-\sqrt{3}\right|\)=\(2-\sqrt{3}\)

k mình nha bn

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

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

=\(2-\sqrt{3}\)( Vì \(2>\sqrt{3}\))

\(\sqrt{a}+\sqrt{b}+\sqrt{c}=3< =>\left(a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\right)=9< =>\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=2\\ \\ \)
Ở đâu có 2 thì thay vào @@
 

Ta có:

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\left(a+b+c\right)+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

\(\Rightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=\frac{3^2-5}{2}=2\)

Ở đâu có 2 thay bằng \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)  là được