tính
a) \(\sqrt{2-\sqrt{3}}\) + \(\sqrt{2+\sqrt{3}}\)
b) \(\sqrt{2+\sqrt{5}}\) - \(\sqrt{2-\sqrt{5}}\)
c) \(\sqrt{3-\sqrt{5}}+\sqrt{7-\sqrt{3}+}+\sqrt{2}\)
help , hix , mk stupit so ai help me với
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19 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}\)
\(=\sqrt{2}\)
c: \(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
12 tháng 10 2021
c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)
TN
1
19 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
13 tháng 5 2022
a.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
b.\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
c.\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}=\left|\sqrt{2}+5\right|-\sqrt{2}=\sqrt{2}+5-\sqrt{2}=5\)
d.\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}=\left|3+\sqrt{5}\right|+\left|\sqrt{5}-6\right|=3+\sqrt{5}+6-\sqrt{5}=9\)
30 tháng 5 2023
a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}\)
b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)
1 tháng 8 2017
b/ \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow B^2=8+2\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow B=\sqrt{5}+1\)
1 tháng 8 2017
a/ \(\sqrt{5\sqrt{3}+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+20\sqrt{3}-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+20\sqrt{3}-20-10\sqrt{3}}\)
\(=\sqrt{15\sqrt{3}-20}\)
a) \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{2}.\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
ok mk giải dk tối qua rồi , dù s cx thanks