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Nguyễn Phương Nhi
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soyeon_Tiểubàng giải
16 tháng 9 2016 lúc 22:19

\(H=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(H=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(H=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(H=\frac{2^{19}.3^9+2^{18}.5.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)

\(H=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+2.3\right)}\)

\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+6\right)}\)

\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)

\(K=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(K=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}\)

\(K=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^{22}}\)

\(K=\frac{2^{22}}{3.2^{23}-5.2^{22}}\)

\(K=\frac{2^{22}}{2^{22}.\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\)

Nguyễn Phương Nhi
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soyeon_Tiểubàng giải
16 tháng 9 2016 lúc 21:51

\(F=\frac{15.3^{11}+4.27^4}{9^7}=\frac{5.3.3^{11}+4.\left(3^3\right)^4}{\left(3^2\right)^7}=\frac{5.3^{12}+4.3^{12}}{3^{14}}=\frac{3^{12}.\left(5+4\right)}{3^{14}}=\frac{3^{12}.9}{3^{12}.9}=1\)

\(G=\frac{5.2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)

\(G=\frac{5.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}{3^2.2^{34}}\)

\(G=\frac{5.2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)

\(G=\frac{5.2^{35}-2^{36}}{3^2.2^{34}}\)

\(G=\frac{2^{35}.\left(5-2\right)}{3^2.2^{34}}\)

\(G=\frac{2^{35}.3}{3^2.2^{34}}=\frac{2}{3}\)

 

Buddy
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HT.Phong (9A5)
18 tháng 8 2023 lúc 18:02

a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)

b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)

c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)

0o0_Đừng_Nhìn_Mình_0o0
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shitbo
7 tháng 12 2018 lúc 19:16

\(a,\frac{4^5.2^{16}}{16^6}=\frac{2^{10}.2^{16}}{2^{24}}=\frac{2^{26}}{2^{24}}=2^2=4\)

\(b,\left(1-\frac{2}{5}\right)^2+|\frac{-3}{5}|+\frac{-7}{10}=\frac{9}{25}+\frac{3}{5}+\frac{-7}{10}=\frac{24}{25}-\frac{7}{10}=\frac{13}{50}\)

c, tt

Nguyễn Viết Trọng
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Ho Bao Ngoc
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Buddy
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HaNa
18 tháng 8 2023 lúc 18:14

a)

$16^{\alpha }+16^{-\alpha } = (4^2)^{\alpha }+(4^2)^{-\alpha } = 4^{2\alpha }+4^{-2\alpha }$

$4^{2\alpha }+4^{-2\alpha } = 4^{2\log_4{\frac{1}{5}}}+4^{-2\log_4{\frac{1}{5}}} = \left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^{-2} = \frac{1}{25}+25 = \frac{26}{25}$

HaNa
18 tháng 8 2023 lúc 18:16

b)

$\left(2^{\alpha }+2^{-\alpha }\right)^2 = \left(\sqrt{4}\right)^{\alpha }+\left(\sqrt{4}\right)^{-\alpha } = 4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}}$

$4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}} = 4^{\frac{\log_4{\frac{1}{5}}}{2}}+4^{-\frac{\log_4{\frac{1}{5}}}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}}+\left(\frac{1}{5}\right)^{-\frac{1}{2}} = \sqrt{\frac{1}{5}}+\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$

Hinamori Amu
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Lightning Farron
26 tháng 10 2016 lúc 18:24

\(M=4\frac{1}{3}-\sqrt{16}+5\sqrt{\frac{4}{9}}-\frac{25}{\left(\sqrt{6}\right)^2}\)

\(=\frac{13}{3}-4+5\cdot\frac{2}{3}-\frac{25}{6}\)

\(=\frac{1}{3}+\frac{10}{3}-\frac{25}{6}\)

\(=\frac{11}{3}-\frac{25}{6}\)

\(=-\frac{1}{2}\)

Kỳ Duyên Tô Nguyễn
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