ai tick mik mik tick lại cho nhg pải giải cho mik cả 3 bài

nói dễ sao ko bay vô làm

1) \(\sqrt{36+12\sqrt{5}}=\sqrt{\left(\sqrt{30}+\sqrt{6}\right)^2}=\sqrt{30}+\sqrt{6}\)

2)\(\sqrt{21-6\sqrt{6}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=\sqrt{18}-\sqrt{3}\)

3)\(\sqrt{6-2\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{9}-1\right)^2}\)

\(=\sqrt{5}-1-\left(\sqrt{9}-1\right)\)

\(=\sqrt{5}-\sqrt{9}\)

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

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

5) \(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

6)\(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

7)\(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}=\sqrt{\left(\sqrt{20}-1\right)^2}+\sqrt{\left(\sqrt{20}+1\right)^2}\)

\(=\sqrt{20}-1+\sqrt{20+1}=2\sqrt{20}\)

a. Ta có:

\(\left(m+1\right)^2\)\(=m^2+2m+1\)

\(\left(m+1\right)^2\ge4m\Leftrightarrow m^2+2m+1\ge4m\)

\(\Leftrightarrow m^2+2m+1-4m\ge0\)

\(\Leftrightarrow m^2-2m+1\ge0\)

\(\Leftrightarrow\left(m-1\right)^2\ge0\) (đúng \(\forall\) m)

Vậy \(\left(m+1\right)^2\ge4m\)

b. \(m^2+n^2+2\ge2\left(m+n\right)\)

\(\Leftrightarrow m^2+1+n^2+1\ge2m+2n\)

Ta có:

\(\left(m^2+1\right)^2\ge4m^2\) \(\Rightarrow m^2+1\ge2m\)

\(\left(n^2+1\right)^2\ge4n^2\Rightarrow n^2+1\ge2n\)