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\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(M^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2\)
\(M^2=\left(\sqrt{4+\sqrt{7}}\right)^2-2.\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}+\left(\sqrt{4-\sqrt{7}}\right)^2\)
\(M^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(M^2=8-2\sqrt{16-7}\)
\(M^2=8-2\sqrt{9}=8-2.3=8-6=2\)
\(M=\frac{+}{ }\sqrt{2}\)
a: Ta có: \(4\sqrt{3a}-3\sqrt{12a}+\dfrac{6\sqrt{a}}{3}-2\sqrt{20a}\)
\(=4\sqrt{3a}-6\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)
\(=-2\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)
a,Ta có : \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)
Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
b, Đặt A = \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)
\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)
Vậy (*) = 0
1:
Ta có: \(\sqrt{2}-\sqrt{6}\)
\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)
\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
Đặt \(N=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Rightarrow N\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=\sqrt{7}+1-\sqrt{7}+1=2\)
\(\Rightarrow N=\sqrt{2}\)
\(\Rightarrow M=N-\sqrt{8}=\sqrt{2}-\sqrt{8}\)