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\(1.\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}=5+\sqrt{7}-\sqrt{7-2\sqrt{7}+1}=5+\sqrt{7}-\sqrt{7}+1=6\)
\(2.\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\sqrt{3}+1=2\)
\(3.VT=\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{11-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{11}+3=3=VP\)
Vậy , đẳng thức được chứng minh .
\(4.VT=\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=\sqrt{36+2.6\sqrt{5}+5}-\sqrt{36-2.6\sqrt{5}+5}=6+\sqrt{5}-6+\sqrt{5}=2\sqrt{5}=VP\)
Vậy , đẳng thức được chứng minh .
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3}+9}=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}=\)
\(\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1=1.\)
a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\)
b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)
c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
\(\sqrt{\left(2\sqrt{2}-3\right)^2}+2\sqrt{2}=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3\)
\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}=\left|\sqrt{10}-3\right|+\left|\sqrt{10}-4\right|\)
\(=\sqrt{10}-3+4-\sqrt{10}=1\)
\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|2-\sqrt{3}\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{\left(6-\sqrt{5}\right)^2}-\sqrt{\left(6+\sqrt{5}\right)^2}\)
\(=6-\sqrt{5}-6-\sqrt{5}=-2\sqrt{5}\)
\(A=\sqrt{49a^2}+3a=7\left|a\right|+3a\)
Nếu \(a\ge0\)thì: \(A=7a+3a=10a\)
Nếu \(a< 0\)thì: \(A=-7a+3a=-4a\)
\(B=3\sqrt{9a^6}-6a^3=9\left|a^3\right|-6a^3\)
Nếu \(a\ge0\)thì: \(B=9a^3-6a^3=3a^3\)
Nếu \(a< 0\)thì: \(B=-9a^3-6a^3=-15a^3\)
Lời giải:
\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{6^2+5-2.6\sqrt{5}}-\sqrt{6^2+5+2.6\sqrt{5}}\)
\(=\sqrt{(6-\sqrt{5})^2}-\sqrt{(6+\sqrt{5})^2}=|6-\sqrt{5}|-|6+\sqrt{5}|\)
\(=(6-\sqrt{5})-(6+\sqrt{5})=-2\sqrt{5}\)