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\(M=\dfrac{5^4\cdot50}{5^3\cdot15}=\dfrac{50}{3}>\dfrac{50}{4}=N\)
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\(4n+1⋮2n-1\)
\(\Leftrightarrow2\left(2n-1\right)+3⋮2n-1\)
\(\Leftrightarrow3⋮2n-1\)
\(\Leftrightarrow2n-1\in\left\{1;3;-1;-3\right\}\)
\(\Leftrightarrow2n\in\left\{2;4;0;-2\right\}\)
\(\Leftrightarrow n\in\left\{1;2;0;-1\right\}\)
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Bài 2:
a: \(\left(6x-39\right):7=3\)
\(\Leftrightarrow6x-39=21\)
hay x=10
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A=\(\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.....\dfrac{99.101}{100.100}\)
A=\(\dfrac{1.2.3.....99}{2.3.4.....100}.\dfrac{3.4.....101}{2.3.4.....100}\)
A=\(\dfrac{1}{100}.\dfrac{101}{2}\)
A=\(\dfrac{101}{200}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\\ =\dfrac{1}{2}.\dfrac{101}{100}=\dfrac{101}{200}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)...\left(1-\dfrac{1}{10000}\right)\\ =\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
(làm như câu a)
\(\dfrac{-2}{3}.\dfrac{15}{-4}-\dfrac{-1}{5}:\dfrac{1}{-5}\)
\(\dfrac{5}{2}-1=\dfrac{3}{2}\)
5/2 - 1
= 3/2