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\(\dfrac{n+4}{n+1}=\dfrac{n+1+3}{n+1}=1+\dfrac{3}{n+1}\)
Để n+4/n+1 nguyên thì 3/n+1 nguyên <=> n+1 thuộc Ư(3)={1;3}
n+1=1 <=> n=0
n+1=3 <=> n=2
Vậy....
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)(ĐPCM)
đề tìm gì bạn
\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\Leftrightarrow\dfrac{1}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)
<=>\(x=\dfrac{6}{5-2y}\)
Để x là số tự nhiên thì 6/5-2y cũng là số tự nhiên <=> \(5-2y\inƯ\left(6\right)\)
Mà 5-2y lẻ nên 5-2y thuộc {1;3}
<=> y thuộc {2;1}
*Với y=2 thì x=6
*Với y=1 thì x=2
Vậy các cặp số tự nhiên (x;y) là (6;2) và (2;1)
a, Vì AB2+AC2=152+202=625 cm
BC2=252=625 cm
=> AB2+AC2=BC2 => tg ABC vuông tại A
b, Ta có AB2+AC2=32 cm
BC2=32 cm
=> AB2+AC2=BC2 => tg ABC vuông tại A
Mà AB=AC=4cm
=> tg ABC vuông cân tại A