\(\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}\)

\(=\sqrt{\frac{x-2\sqrt{2x-4}}{2}}\)

\(=\sqrt{\frac{x}{2}-\frac{2\sqrt{2x-4}}{2}}\)

\(=\sqrt{\frac{x}{2}-\sqrt{2x-4}}\)

\(=\sqrt{\frac{x}{2}-\sqrt{2x-4}}\)

Bn có làm đc bài 1 ko

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

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

= \(5\sqrt{2}\)- 1

#mã mã#

\(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2\left(\sqrt[3]{4}-\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\left(\sqrt[3]{4^2}+\sqrt[3]{4}.\sqrt[3]{2}+\sqrt[3]{2^2}\right)}\)
\(=\frac{2\left(\sqrt[3]{4}-\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}\right)^3-\left(\sqrt[3]{2}\right)^3}=\sqrt[3]{4}-\sqrt[3]{2}\)

\(y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{2\left(\sqrt[3]{4}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}+\sqrt[3]{2}\right)\left(\sqrt[3]{4^2}-\sqrt[3]{4}.\sqrt[3]{2}+\sqrt[3]{2^2}\right)}\)

\(=\frac{6\left(\sqrt[3]{4}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}\right)^3+\left(\sqrt[3]{2}\right)^3}=\sqrt[3]{4}+\sqrt[3]{2}\)

\(P=\frac{xy}{x+y}=\frac{\sqrt[3]{4^2}-\sqrt[3]{2^2}}{2\sqrt[3]{4}}=\frac{\sqrt[3]{4}-1}{2}\)

=>(x-\(\sqrt{5}\))²

=>(x-\(\sqrt{5}\)) (x-\(\sqrt{5}\))

Do m, n cùng dấu, m, n khác 0 nên m, n cùng âm hoặc cùng dương, mà nếu m, n cùng âm thì \(\frac{1}{2m}+\frac{1}{n}< 0< \frac{1}{3}\)

trái với gt \(\Rightarrow\) m, n cùng dương 

\(\frac{1}{3}=\frac{1}{2m}+\frac{1}{n}\ge2\sqrt{\frac{1}{2mn}}\)\(\Leftrightarrow\)\(\frac{1}{2mn}\le\frac{1}{36}\)\(\Leftrightarrow\)\(mn\ge18\)\(\Rightarrow\)\(B\ge18\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\frac{1}{2m}=\frac{1}{n}\\\frac{1}{2m}+\frac{1}{n}=\frac{1}{3}\end{cases}\Leftrightarrow\hept{\begin{cases}m=3\\n=6\end{cases}}}\)

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

Tách: \(\sqrt{15}=\sqrt{15}.1\) mà \(\left(\sqrt{15}\right)^2+1^2=16\ne8\)loại

\(\sqrt{15}=\sqrt{3}.\sqrt{5}\), \(\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2=8\)nhận 

\(\sqrt{8-2\sqrt{15}}=\sqrt{3-2\sqrt{3}.\sqrt{5}+5}=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=\left|\sqrt{3}-\sqrt{5}\right|\)

\(=\sqrt{5}-\sqrt{3}\)

Th1 b <0

ab² \(\sqrt{\frac{3}{a^2b^2}}\)=ab² . \(\frac{\sqrt{3}}{ab}\)= \(-b\sqrt{3}\)

th2 b>0

ab² \(\sqrt{\frac{3}{a^2b^2}}\)= ab² . . \(\frac{-\sqrt{3}}{ab}\)= -b\(\sqrt{3}\)

#mã mã#