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S=5+5^2+5^3+5^4+...5^99
=> 5S=5^2+5^3+5^4+...5^100
=> 5S-S=4S=(5^2+5^3+5^4+...5^100)-(5+5^2+5^3+5^4+...5^99)
=> 4S = 5100-5
=> S=(5100-5)/4
S=5*5^2*5^3*5^4*...5^99
=> S=51+2+3+...+99
=> S=5((99+1).99):2
=> S=54950
(3x-2)+16=125+12
(3x-2)+16=137
3x-2=121
3x=123
x=41
5s=5^2+5^3+5^4+5^5+......+5^100
5s-s=5^100-5
4s=5^100-5
s=(5^100-5):4
kick nhé
\(\left(3x-2\right)^2+4^2=5^3+3.2^2\\ \Rightarrow\left(3x-2\right)^2+16=125+12\\ \Rightarrow\left(3x-2\right)^2=121\\ \Rightarrow3x-2=11\\ \Rightarrow x=\frac{13}{3}\)
S= \(5+5^2+5^3+.....+5^{99}\\ \Rightarrow5S=5^2+5^3+5^4+.....+5^{100}\\ \Rightarrow4S=5^{100}-5\\ \Rightarrow\frac{5^{100}-5}{4}\)
S=\(5.5^2.5^3.5^4.........5^{99}=5^{1+2+3+4+....+99}=5^{4950}\)
S=5+52+53+54+...+599
5S=52+53+54+...+599
5S-S=(52-52)+(53-53)+...+(599-599)+5100+5
S=(5100+5):4
Bài 1 :
a, \(\frac{3}{4}:x=\frac{5}{12}\)
\(x=\frac{3}{4}:\frac{5}{12}\)
\(x=\frac{9}{5}\)
b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)
\(\frac{3}{2}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{3}{2}\)
\(x=\frac{5}{6}\)
Bài 2 :
\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)
\(A=\frac{-3}{5}+\frac{-2}{5}-99\)
\(A=\left(-1\right)-99\)
\(A=-100\)
\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)
\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)
\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)
\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)
\(B=1+\frac{13}{5}\)
\(B=\frac{18}{5}\)
Các bn ơi giải hộ mik với. Ai giải đầu mik sẽ k cho. Cảm ơn các bn nhiều nha.
Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
=>\(3S=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\ldots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
=>3S+S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
=>4S=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)
=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)
=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)
=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)
=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)
Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)
=>\(S<\frac{3}{16}\)
mà 3/16<3/15=1/5
nên S<1/5