Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{900}\right)\)
\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)
\(=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{3^2-1}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{30^2-1}{30^2}\right)\)
\(=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{29.31}{30^2}\right)\)
\(=\left(\dfrac{1.2.3...29}{2.3.4...30}\right).\left(\dfrac{3.4.5...31}{2.3.4...30}\right)=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)
Do đó:
\(M=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)
\(=\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}.\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}.\dfrac{\left(4-1\right)\left(4+1\right)}{4^2}...\dfrac{\left(30-1\right)\left(30+1\right)}{30^2}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{29.31}{30^2}=\dfrac{1.2.3...29}{2.3.4...30}.\dfrac{3.4.5...31}{2.3.4...30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
D=1+.....+4^11chia het cho 5
D=(1+4)+(4^2+4^3)+......+(4^10+4^11)chia het cho 5
D=(1+4)+4^2(1+4)+....+4^10(1+4)chia het cho 5
D=5+4^2.5+....+4^10.5chia het cho 5
D=5(4^2+4^4+....+4^10)chia het cho 5
suy ra Dchia het cho 5 (do 5 chia het cho 5)
vậy Dchia het cho 5
Olm chào em, dưới đây là chú giải cho câu hỏi của em
Nếu p = 3k + 2 ta có:
2p\(^2\) + 1
= 2(3k + 2)\(^2\) + 1
= 2.(9k\(^2\) + 12k + 4) + 1
= 18k\(^2\) + 24k + 8 + 1
= 18k\(^2\) + 24k + (8 + 1)
= 18k\(^2\) + 24k + 9
= 3.(6k\(^2\) + 8k + 3) ⋮ 3
Bài 2 :
Ta có:
A) | 2 + 3x | = | 4x - 3 |
<=> 2 + 3x = 4x - 3
<=> 3x - 4x = -3 - 2
=> -x = -5
=> x= 5
l\(\frac{3}{4}.x-\frac{1}{2}\)l=\(\frac{1}{4}\)
=>\(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\) hoặc \(\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\)
*nếu \(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\)
=>\(\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{1}{4}+\frac{2}{4}=\frac{3}{4}\)
=>\(x=\frac{3}{4}:\frac{3}{4}=1\)
*nếu \(\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\)
=>\(\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=-\frac{1}{4}+\frac{2}{4}=\frac{1}{4}\)
=>\(x=\frac{1}{4}:\frac{3}{4}=\frac{1}{4}.\frac{4}{3}=\frac{1}{3}\)
vậy \(x\in\left\{\frac{1}{3};1\right\}\)
I Am A Good Boy_Wang Jun Kai Ahihi, M lười ghê ý