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Đặt \(A=10+3^2+...+3^{99}\)
Đặt \(B=3^2+3^3+...+3^{99}\)
\(3B=3^3+3^4+...+3^{100}\)
\(3B-B=\left(3^3+3^4+...+3^{100}\right)-\left(3^2+3^3+...+3^{99}\right)\)
\(2B=3^{100}-3^2\)
\(B=\frac{3^{100}-3^2}{2}\)
\(\Rightarrow A=10+\frac{3^{100}-3^2}{2}\)
\(\Rightarrow A=\frac{20+3^{100}-3^2}{2}\)
\(\Rightarrow A=\frac{3^{100}+11}{2}\)

18^3 : 9^3 = 5832 : 729 = 8
125^3 : 25^3 = (5^3)^3 : (5^2)^3 = 5^9 : 5^6 = 5^3 = 125
có quy luật hay lắm
\(18^3:9^3=\left(18:9\right)^3=2^3=8\)
\(125^3:25^3=\hept{\begin{cases}\left(5^3\right)^3:\left(5^2\right)^3=5^9:5^6=5^3=125\\\left(5^3\right)^3:\left(5^2\right)^3=\left(5^3:5^2\right)^3=5^3=125\end{cases}}\)chọn cách nào thì tùy bạn
\(\left(10^3+10^4+125^3\right):5^3=\left[10^3+10^3.10+\left(5^2\right)^3\right]:5^3\)
\(=\left(10^3.11+5^6\right):5^3\)
\(=10^3.11:5^3+5^6:5^3\)
\(=\left(10^3:5^3\right).11+5^3\)
\(=2^3.11+5^3\)
\(=88+125=213\)
\(\left(2^{43}+2^4\right):\left(2^{39}+1\right)=\)tương tự mà làm

\(1:\frac{99}{100}:\frac{98}{99}:\frac{97}{98}:.........:\frac{2}{3}:\frac{1}{2}\)
\(=1.\frac{100}{99}.\frac{99}{98}.\frac{98}{97}......\frac{3}{2}.\frac{2}{1}\)
\(=\frac{1.100.99.98....3.2}{99.98.97......2.1}\)
\(=100\)

\(4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2009\)
\(=4.\left\{9.\left[\left(25^{ }+8\right):11\right]-26\right\}+2009\)
\(=4.\left\{9.\left[33:11\right]-26\right\}+2009\)
\(=4.\left\{9.3-26\right\}+2009\) \(=4.\left(27-26\right)+2009\)
\(=4.1+2009=2013\)

hỗn số thì tử phải lớn hơn mẫu thì chia tử với mẫu thì mới ra hỗn số

a)=\(\frac{7}{30}\)-\(\frac{17}{12}\)=\(\frac{-71}{60}\) b) = \(\frac{29}{20}\)+\(\frac{-31}{30}\)+\(\frac{-2}{3}=\frac{-1}{4}\)
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