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a) \(M=\sqrt{3-2\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2.\sqrt{2}.2+4}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+2\right)^2}=\left|\sqrt{2}-1\right|+\sqrt{2}+2=\sqrt{2}-1+\sqrt{2}+2=2\sqrt{2}+1\)
b) \(N=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\left|\sqrt{3}-1\right|}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{2}.\sqrt{3}=\sqrt{6}\)
Hướng dẫn trả lời:
M=√3−2√2−√6+4√2=√(√2)2−2√2.1+12−√(2)2+2√2+(√2)2=√(√2−1)3−√(2+√2)2=∣∣√2−1∣∣−∣∣2+√2∣∣=√2−1−2−√2=−3M=3−22−6+42=(2)2−22.1+12−(2)2+22+(2)2=(2−1)3−(2+2)2=|2−1|−|2+2|=2−1−2−2=−3
N=√2+√3+√2−√3⇒N2=(√2+√3+√2−√3)2=2+√3+2√(2+√3)(2−√3)+2−√3=4+2√4−3=6N=2+3+2−3⇒N2=(2+3+2−3)2=2+3+2(2+3)(2−3)+2−3=4+24−3=6
Vì N > 0 nên N2 = 6 ⇒ N = √6. Vậy
\(N=\sqrt{6+2\sqrt{2}.\sqrt{3+\sqrt{4-2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}\)\(=\sqrt{6+2\sqrt{2}.\sqrt{3+\sqrt{3}-1}=}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\) \(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
a) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{1}{2\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2\sqrt{3}}{12}+\frac{2\sqrt{3}}{6}-\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}}{12}+\frac{4\sqrt{3}}{12}-\frac{12-4\sqrt{3}}{12}=\frac{-12+10\sqrt{3}}{12}=\frac{-6+5\sqrt{3}}{6}\)