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\(\dfrac{-178}{179}>-1>\dfrac{-191}{189}\\ \dfrac{127}{129}=1-\dfrac{2}{129};\dfrac{871}{873}=1-\dfrac{2}{873}\\ \dfrac{2}{129}>\dfrac{2}{873}\left(129< 873\right)\Leftrightarrow1-\dfrac{2}{129}< 1-\dfrac{2}{873}\Leftrightarrow\dfrac{127}{129}< \dfrac{871}{873}\)
12345 x 1256 - 1238 - 129 = 15505320 - 1238 - 129 = 15504082 - 129 = 15503953
A=1/30 + 1/42 + 1/56 + 1/72 + ... + 1/210
A=1/5.6 + 1/6.7 +1/7.8 + 1/8.9 + ... + 1/14.15
A=1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 +...+ 1/14 - 1/15
A=1/5-1/15
A=2/15
a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)