Trả lời:

\(\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right).\left(\sqrt{18}-\sqrt{20}+2\sqrt{2}\right)\)

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

\(=\left(5\sqrt{2}-\sqrt{5}\right).\left(5\sqrt{2}-2\sqrt{5}\right)\)

\(=50-10\sqrt{10}-5\sqrt{10}+10\)

\(=60-15\sqrt{10}\)

Trả lời:

\(G=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\frac{2}{\sqrt{2}}.G=\frac{2}{\sqrt{2}}.\left(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\right)\)

\(\sqrt{2}.G=\frac{2.\left(2+\sqrt{3}\right)}{\sqrt{2}.\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{2.\left(2-\sqrt{3}\right)}{\sqrt{2}.\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)

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

\(\sqrt{2}.G=\frac{3+2\sqrt{3}+1}{2+\sqrt{3+2\sqrt{3}+1}}+\frac{3-2\sqrt{3}+1}{2-\sqrt{3-2\sqrt{3}+1}}\)

\(\sqrt{2}.G=\frac{\left(\sqrt{3}+1\right)^2}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\left(\sqrt{3}-1\right)^2}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(\sqrt{2}.G=\frac{\left(\sqrt{3}+1\right)^2}{2+\sqrt{3}+1}+\frac{\left(\sqrt{3}-1\right)^2}{2-\sqrt{3}+1}\)

\(\sqrt{2}.G=\frac{\left(\sqrt{3}+1\right)^2}{3+\sqrt{3}}+\frac{\left(\sqrt{3}-1\right)^2}{3-\sqrt{3}}\)

\(\sqrt{2}.G=\frac{\left(\sqrt{3}+1\right)^2}{\sqrt{3}.\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{\sqrt{3}.\left(\sqrt{3}-1\right)}\)

\(\sqrt{2}.G=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}\)

\(\sqrt{2}.G=\frac{2\sqrt{3}}{\sqrt{3}}\)

\(\sqrt{2}.G=2\)

\(G=\sqrt{2}\)

Học tốt 

Vì \(x\ne0,y\ne0\) nên điều kiện đã cho tương đương với \(\frac{x}{y^2}+\frac{y}{x^2}=2\Rightarrow\frac{x^2}{y^4}+\frac{y^2}{x^4}+\frac{2}{xy}=4\Leftrightarrow4\left(1-\frac{1}{xy}\right)=\frac{x^2}{y^4}+\frac{y^2}{x^4}-\frac{2}{xy}=\left(\frac{x}{y^2}-\frac{y}{x^2}\right)^2\)

\(\Rightarrow\sqrt{1-\frac{1}{xy}}=\frac{1}{2}\left|\frac{x}{y^2}-\frac{y}{x^2}\right|\)

Trả lời:

\(\sqrt{x^2-4x+4}=x^2-mx+2m-4\)\(\left(ĐK:x\ge2\right)\)

\(\sqrt{\left(x-2\right)^2}=x^2-mx+2m-4\)

\(x-2=x^2-mx+2m-4\)

\(x^2-mx+2m-x-2=0\)

\(x^2-\left(m+1\right).x+2m-2=0\)

\(\Delta=\left[-\left(m+1\right)\right]^2-4.\left(2m-2\right)\)

\(=m^2+2m+1-8m+8\)

\(=m^2-6m+9\)

\(\Rightarrow\left(m-3\right)^2\ge0\)

\(\Rightarrow m-3=0\)

\(\Rightarrow m=3\)

Thay m=3 vào phương trình ta có:

\(x^2-\left(3+1\right).x+2.3-2=0\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy \(x=2\Leftrightarrow m=3\)

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

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

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

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)