√(3 - √(5)) × (3 + √(5)) / √10+√2
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\(\frac{4}{5}+\frac{3}{15}=\frac{4}{5}+\frac{1}{5}=\frac{4+1}{5}=\frac{5}{5}=1\)
\(\frac{2}{3}+\frac{32}{24}=\frac{2}{3}+\frac{4}{3}=\frac{2+4}{3}=\frac{6}{3}=2\)
\(\frac{5}{6}+\frac{15}{18}=\frac{5}{6}+\frac{5}{6}=\frac{5+5}{6}=\frac{10}{6}\)
HT
A = 5 + 5 ^ 2 + 5 ^ 3 + ... + 5 ^ 50
5 A = 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + ... + 5 ^ 51
5 A - A = ( 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + ... + 5 ^ 51 )
- ( 5 + 5 ^ 2 + 5 ^ 3 + ... + 5 ^ 50 )
4 A = 5 ^ 51 - 5
A = \(\frac{5^{51}-5}{4}\)
A=5^1+5^21+5^3+...+5^50
5^1A=5(5^1+5^2+5^3+..+5^50)
5A=5^2+5^3+..+5^50+5^51
5A-A=(5^2+5^3+..+5^50+5^51)-(5^1+5^2+5^3+..+5^50)
4A=5^51-5^1
A=(5^51-5^1):4
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{25-3}=\frac{8}{22}=\frac{4}{11}\)
\(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{5-9}{14}=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-2}{7}=\frac{-22}{77}\)
\(\frac{4}{11}=\frac{28}{77}\)
Ta có: \(\dfrac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)^2}{4\cdot\left(\sqrt{5}+1\right)}\)
\(=\dfrac{4}{4}=1\)