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\(\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(4-\sqrt{5}\right)^2}\)
\(=\sqrt{\left(\sqrt{5-1}^2\right)}\)
\(=|4-\sqrt{5}|-|\sqrt{5}-1|\)
\(=4-\sqrt{5}-\sqrt{5}+1=5-2\sqrt{5}\)
P/s: Khôg chắc lắm ạ
\(=\frac{-40\sqrt{3}+30\sqrt{2}}{-4\sqrt{3}+3\sqrt{2}}=10\)
rút gọn
C=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(C=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
`C=(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}`
`C=(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10})\sqrt{4-\sqrt{15}}`
`C=(\sqrt{10}+\sqrt{6})\sqrt{4-\sqrt{15}}`
`C=\sqrt{(\sqrt{10}+\sqrt{6})^2 .(4-\sqrt{15})}`
`C=\sqrt{(10+6+2\sqrt{60})(4-\sqrt{15})}`
`C=\sqrt{(16+4\sqrt{15})(4-\sqrt{15})}`
`C=\sqrt{64-16\sqrt{15}+16\sqrt{15}-60}`
`C=\sqrt{4}=2`
a) \(\sqrt{5+\sqrt{21}}-\sqrt{6-\sqrt{35}}\) = \(\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}-\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}\) = \(\dfrac{\sqrt{7}+\sqrt{3}-\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{2}}\)
= \(\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)
câu b) hình như đề sai