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Ta có :
\(D=\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-..........-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\)
\(5D=1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+..........+\dfrac{1}{5^{100}}\)
\(5D+D=\left(1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+.........+\dfrac{1}{5^{100}}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5^2}+..............-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\right)\)\(6D=1-\dfrac{1}{5^{101}}\)
\(D=\dfrac{1-\dfrac{1}{5^{101}}}{6}\)
2P=\(\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{100}}\)
2P=\(1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
2P-P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)\(\)
\(2P-P=1-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}}{2^{100}}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}-1}{2^{100}}\)
1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3
=>x=-16/3:7/3=-7/16
2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35
=>|x-2|=129/35
=>x-2=129/35 hoặc x-2=-129/35
=>x=199/35 hoặc x=-59/35
Đặt N=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+......+\dfrac{1}{5^{100}}\)
5N=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+..........+\dfrac{1}{5^{99}}\)
5N-N= \(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.............+\dfrac{1}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+..........+\dfrac{1}{5^{100}}\right)\)
4N=1-\(\dfrac{1}{5^{100}}\) =\(\dfrac{5^{100}-1}{5^{100}}\)
N=\(\dfrac{5^{100}-1}{4.5^{100}}\)
Thay N vào D ,ta có
D= 4.5\(^{100}\).(\(\dfrac{5^{100}-1}{4.5^{100}}\) )+1
D=5\(^{100}\)
Vậy D =5\(^{100}\)
cau 1
de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat
suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong
suy ra 4a-23=1
suy ra 4a=1+23=24
suy ra a=24 chia 4=6
vay de a nho nhat thi a=6
Ta có : M . N = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{101}\)
Vậy M . N = \(\dfrac{1}{101}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
Đặt `B=1/5+1/5^{2}+1/5^{3}+...+1/5^{101}`
`<=>5B=1+1/5+1/5^{2}+...+1/5^{100}`
`<=>5B-B=(1+1/5+1/5^{2}+...+1/5^{100})-(1/5+1/5^{2}+...+1/5^{101})`
`<=>5B-B=1+1/5+1/5^{2}+...+1/5^{100}-1/5-1/5^{2}-...-1/5^{101}`
`<=>4B=1-1/5^{101}`
`<=>B=(1-1/5^{101})/4`
`@Shả`
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{101}}\)
\(5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{100}}\)
\(5A-A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{100}}-\dfrac{1}{5}-\dfrac{1}{5^2}-...-\dfrac{1}{5^{101}}=1-\dfrac{1}{5^{101}}\Rightarrow A=\dfrac{1-\dfrac{1}{5^{101}}}{4}\)