5 + 5.9+5.9^2+5.9^3+.....+5.9^99
RÚT GỌN
\(A=_{100}+5+5.9+5.9^2+5.9^3+...+5.9^{99}\)
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Tính:
a) \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
b) T=\(\dfrac{5^{16}.27^7}{125^5.9^{11}}\)
a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
Cho E=2^15.0,5^5+3.2^10/2^11.2^3-2^15:2^3 Cho F=2.6^7+6^8/2^5.9^4-2.3^7+3^8 Cho G=4^5.9^4-2.6^9/2^10.3^8+6^8.20
Bài 1 : Tính
a , `5 + 5^2 + 5^3 + ... + 5^99`
b, `\frac{1}{1.5} + \frac{1}{5.9} + ... + \frac{1}{17.21}`
a, Giải:
Đặt A = \(5+5^2+5^3+...+5^{99}\)
5A = \(5^2+5^3+...+5^{100}\)
5A - A = ( \(5^2+5^3+...+5^{100}\)) - ( \(5+5^2+5^3+...+5^{99}\))
4A = \(5^{100}-5\)
A = \(\frac{5^{100}-5}{4}\)
b, Giải:
Đặt B = \(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{17.21}\)
4B = 4. (\(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{17.21}\))
4B = \(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{17.21}\)
4B = \(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\)
4B = 1 - \(\frac{1}{21}\)
4B = 20/21
B = \(\frac{\frac{20}{21}}{4}\) = 5/21
Tính nhanh:
a) 3/5+3/5.9+3/9.13+...+3/97.101
b) (1+1/2).(1+1/3).(1+1/4)...(1+1/99)
GIÚP MÌNH VỚI Ạ.CẢM ƠN MỌI NGƯỜI!
\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)
\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)
\(=\dfrac{75}{101}\)
\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Tính nhanh:
a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\times\dfrac{100}{101}\)
= \(\dfrac{75}{101}\)
b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)
\(=\dfrac{100}{2}\)
\(=50\)
3^5.9^2 / 3^3
5.9^2:8=2x
\(2x=5.9^2:8\)
\(2x=5.81:8\)
\(16x=405\)
\(x=\dfrac{405}{16}\)
\(5\cdot9^2:8=2x\\\Rightarrow 2x=5\cdot81:8\\\Rightarrow2x=405:8\\\Rightarrow2x=\dfrac{405}{8}\\\Rightarrow x=\dfrac{405}{16}\\Vậy:x=\dfrac{405}{16}\)
2.x-49=5.9
`2.x-49=5.9`
`=> 2.x-49=45`
`=>2.x=45+49`
`=>2.x=94`
`=>x=94:2`
`=>x=47`
2 . x - 49 = 45
2 . x = 49 + 45
2 . x = 94
x = 94 : 2
x = 47