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\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
Lời giải:
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)
\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)
\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)
b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)
\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)
\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)
\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)
\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)
Ta có
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\) nên
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)
\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)
Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)
do vế trái luôn luôn lớn hơn hoặc =0
=> vế phải cx luôn luôn lớn hơn hoặc =0
=> bỏ giá trị tuyệt đối =100x
có 99x + ........... = 100x
trừ là ra nha bn
ta có:
|x+1/1.2|+|x+1/2.3|+...+|x+1/99.100|=100x
=>|x+1/1.2+x+1/2.3+...+x+1/99.100|=100x
<=>|(x+x+x+...+x)+1/1.2+1/2.3+....1/99.100|=100x
<=>|x.99+1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100|=100x
<=>|x.99+1-1/100|=100x
<=>|99x+99/100|=100x
Có 2 trường hợp
TH1
99x+99/100=100x
=>100x-99x=99/100
<=>x=99/100
=>x=99/100
TH2:
99x+99/100=-100x
-100x-99x=99/100
<=>-199x=99/100
<=>x=99/-19900( loại vì |99x+99/100| là số dương nên 100x là số dương mà x là sô âm nên 100x là số âm)
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)
1. Theo bài ra, ta có:
a + b = ab
⇒ a = ab - b
⇒ a = b ( a - 1 )
⇒ \(\dfrac{a}{b}\) = a - 1
Vậy \(\dfrac{a}{b}\) = a - 1 ( Điều phải chứng minh )
a) A = 1/(1.2) + 1/(2.3) + ... + 1/[n(n + 1)]
= 1 - 1/2 + 1/2 - 1/3 + 1/n - 1/(n + 1)
= 1 - 1/(n + 1)
b) Do n ∈ ℕ
⇒ n + 1 > 0
⇒ 1/(n + 1) > 0
⇒ 1 - 1/(n + 1) < 1
Vậy A < 1