Ta có:

\(4a^2+a\sqrt{2}-\sqrt{2}=0\)

\(\Leftrightarrow2\sqrt{2}a^2+a-1=0\)

\(\Leftrightarrow a+1=2-2\sqrt{2}a^2\) thế vô ta được

\(\frac{a+1}{\sqrt{a^4+a+1}-a^2}=\frac{2-2\sqrt{2}a^2}{\sqrt{a^4+2-2\sqrt{2}a^2}-a^2}\)

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

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

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

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

\(A=\sqrt{6}\)

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

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

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

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

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

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:

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

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

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

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