\(\sqrt{8-2\sqrt{15}}\)
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4: \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\)
\(=2\sqrt{3}\)
4) \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{5}+\sqrt{3}-\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}\)
5) \(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\\ =\sqrt{\sqrt{5^2}+2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\sqrt{5^2}-2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left|\sqrt{5}+\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\\ =2\sqrt{3}\)
\(b,\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\\ =\sqrt{\sqrt{2^2}+2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}+\sqrt{\sqrt{2^2}-2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\\ =\left|\sqrt{2}+\sqrt{3}\right|+\left|\sqrt{2}-\sqrt{3}\right|\\ =\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
a) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5-2\cdot\sqrt{5\cdot3}+3}-\sqrt{5+2\cdot\sqrt{5\cdot3}+1}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(=-2\sqrt{3}\)
b. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{3-2\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)
\(\sqrt{8-2\sqrt{15}}+\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)
\(=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
2: \(\dfrac{\sqrt{108}}{\sqrt{3}}=6\)
13: \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
14: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
=2
12.
\(\dfrac{\sqrt{108}}{\sqrt{3}}=\dfrac{\sqrt{36}.\sqrt{3}}{\sqrt{3}}=\sqrt{36}=6\)
13.
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{5}\right|-\left|2\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{5-2\sqrt{5}.\sqrt{3}+3}-\sqrt{5+2.\sqrt{5}.\sqrt{3}+3}=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\)
Ta có: \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|-\left|\sqrt{3}+\sqrt{5}\right|\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}\)
\(=-2\sqrt{3}\)
\(\sqrt{A^2}=\left|A\right|\)
Tách: \(\sqrt{15}=\sqrt{15}.1\) mà \(\left(\sqrt{15}\right)^2+1^2=16\ne8\)loại
\(\sqrt{15}=\sqrt{3}.\sqrt{5}\), \(\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2=8\)nhận
\(\sqrt{8-2\sqrt{15}}=\sqrt{3-2\sqrt{3}.\sqrt{5}+5}=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=\left|\sqrt{3}-\sqrt{5}\right|\)
\(=\sqrt{5}-\sqrt{3}\)