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Bài 1 tào lao quá, đương nhiên A > B
Bài 2:
Cách 1: A={7;9}
Cách 2: A={x\(\in\)N|x\(\notin\)Ư(2)|6<x<11|
A có 2 phần tử
Làm từ từ chút
A = 1/99 - 1/99.98 - 1/98.97 - ............... - 1/3.2 - 1/2.1
\(A=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(B=\frac{1}{99.98}+\frac{1}{97.87}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(B=1-\frac{1}{99}\)
\(B=\frac{98}{99}\)
\(\Rightarrow A=\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}\)\(+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\frac{1}{3}\)
\(=\frac{7}{12}\)
a, \(\frac{25.5^3.1}{25.5^2}=\frac{5^2.5^3.1}{5^2.5^2}=\frac{5^5}{5^4}=5\)
b, \(5^2.3^5.\left(\frac{3}{5}\right)^2=5^2.3^5.\frac{9}{25}=5^2.\frac{9}{25}.3^5=9.3^5=3^2.3^5=3^7\)
c, \(\left(\frac{1}{7}\right)^2\cdot\frac{1}{7}\cdot\frac{49}{2}=\frac{1}{49}\cdot\frac{1}{7}\cdot\frac{49}{2}=\frac{1}{49}\cdot\frac{49}{2}\cdot\frac{1}{7}=\frac{1}{2}\cdot\frac{1}{7}=\frac{1}{14}\)
Bài 2:
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{999\cdot1000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
=1-1/1000
=999/1000
a.2.3/5:2.1/4+5/2
=2.3.4/5.2.1+5/2
=3.4/5+5.2
=12/5+5/2=24/10+25/10=49/10
b.16/6-2.1/6:1.1/7
=16/6-2/6.1/7
=16/6-2/42
=55/21
a)2×3/5÷2×1/4+5/2
=6/5÷1/2+5/2
=12/5+5/2
=49/10
b)16/6-2×1/6÷1×1/7
=8/3-1/3÷1/7
=8/3-7/3
=1/3