Question

[Tsukino Usagi]

a/ A = \(\frac{5^4x20^4}{25^5x4^5}\).

b/ B = \(\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}\).

c/ C = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\).

d/ D = \(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\).

Mình đang cần gấp, ai làm nhanh và đúng mik sẽ tik cho 3 cái 3 hôm nha!

( Dấu " x " ở câu a với câu b là dấu nhân đó).

Answer 1

[Hoàng Ninh]

a) \(A=\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(2^2.5\right)^4}{5^{2^5}.\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{\left(5^{10}:5^8\right).\left(2^{10}:2^8\right)}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}\)

b) \(B=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)\(=\frac{2^3+2^3}{1}=\frac{8+8}{1}=16\)

c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\)

\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{99}}\)

\(\Rightarrow2C-C=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\right)\)

\(\Rightarrow C=1-\frac{1}{2^{100}}\)

d) \(D=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{100}}\)

\(\Rightarrow5D=5+1+\frac{1}{5^2}+\frac{1}{5^3}+...........+\frac{1}{5^{101}}\)

\(\Rightarrow5D-D=\left(5+1+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{101}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..........+\frac{1}{5^{100}}\right)\)

\(\Rightarrow4D=5-\frac{1}{5^{101}}\)

\(\Rightarrow D=\frac{5-\frac{1}{5^{101}}}{4}\)

Answer 2

[I don't need you]

a) \(A=\frac{5^4x20^4}{25^5x4^5}=\frac{5^4x\left(2^2x5\right)^4}{\left(5^2\right)^5x\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2x2^2}=\frac{1}{25.4}=\frac{1}{100}\)

b) \(B=\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}=\frac{2^{13}.5^7.\left(2^{17}+5^{20}\right)}{2^{10}.5^7.\left(2^{17}+5^{20}\right)}=2^3=8\)

c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2C-C=1-\frac{1}{2^{100}}\)

\(C=1-\frac{1}{2^{100}}\)

phần d bn lm tương tự như phần c nha!