Vì ab = cd nên \(\dfrac{a}{c}=\dfrac{d}{b}\)

Đặt \(\dfrac{a}{c}=\dfrac{d}{b}=k\) (k > 0)

=> a = ck ; d = bk

Khi đó P = aⁿ + bⁿ + cⁿ + dⁿ

= (ck)ⁿ + bⁿ + cⁿ + (bk)ⁿ

= cⁿ.kⁿ + cⁿ + bⁿ + bⁿ.kⁿ

= cⁿ(kⁿ + 1) + bⁿ(kⁿ + 1)

= (cⁿ + bⁿ).(kⁿ + 1) 

Dễ thấy cⁿ + bⁿ > 1 ; kⁿ + 1 > 1

=> P là hợp số 

Hellp me !!!!!!!!!!!!!!!!!

ko có thời giân để hỏi đâu8

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`1/3 - 4/9 xx x =3/5`

`=> 4/9 xx x =1/3-3/5`

`=> 4/9 xx x = 5/15 -9/15`

`=> 4/9 xx x = -4/15`

`=>x= -4/15 : 4/9`

`=>x=-4/15 xx 9/4`

`=>x=-3/5`

S = 1 + 1/1! + 1/2! + 1/3! + ... + 1/2001

=2+1/2+......+1/2001

có

1/2=1.2

1/3<1/2.3

......

1/2001<1/2000.2001

1/2! + 1/3! + ... + 1/2001<1/1.2+1/1.3+.....+1/2000.2001

1/2! + 1/3! + ... + 1/2001<1-1/2+1/2-1/3+....+1/2000-1/2001

1/2! + 1/3! + ... + 1/2001<1-1/2001<1

1/2! + 1/3! + ... + 1/2001<1

vậy:1/2! + 1/3! + ... + 1/2001<3     nhớ gửi coin nhé! chúc bạn làm đúng :))

Ta có:
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(...\)
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)     \(\left(1\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{3}-\dfrac{1}{100}\)\(< \dfrac{1}{3}\)     \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)