Tính:
\(\left(\frac{2}{5}.\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)
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\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
Đặt \(a=\frac{1-\sqrt{5}}{2},b=\frac{1+\sqrt{5}}{2}\)
Ta có \(a+b=1,a-b=-\sqrt{5},ab=-1\)
Ta sẽ tính từ từ. Cụ thể
\(a^2+b^2=\left(a+b\right)^2-2ab=3\)
\(a^2-b^2=\left(a+b\right)\left(a-b\right)=-\sqrt{5}\)
\(a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=7\)
\(a^4-b^4=\left(a^2+b^2\right)\left(a^2-b^2\right)=-3\sqrt{5}\)
\(a^8+b^8=\left(a^4+b^4\right)^2-2\left(ab\right)^4=47\)
\(a^8-b^8=\left(a^4+b^4\right)\left(a^4-b^4\right)=-21\sqrt{5}\)
\(a^{16}-b^{16}=\left(a^8+b^8\right)\left(a^8-b^8\right)=-987\sqrt{5}\)
\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)
\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)
\(=\frac{5}{2}-\frac{8}{27}-8\)
\(=\frac{-313}{54}\)
\(\left(\frac{2}{5}\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}=\left(\frac{2}{5}.\sqrt{4^2}+2\sqrt{\frac{4^2}{5^2}}\right):\frac{2}{\sqrt{4^2}}\)
\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right).2=\left(\frac{8}{5}+\frac{8}{5}\right).2=\frac{32}{5}\)
\(\left(\frac{2}{5}.\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)
\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right):2.\frac{1}{4}\)
\(=\left(\frac{8}{5}+\frac{8}{5}\right):\frac{1}{2}\)
\(=\frac{16}{5}:\frac{1}{2}\)
\(=\frac{32}{5}\)