Cho A=(1/22-1)(1/32-1)(1/42-1)......(1/20132-1)(1/20142-1) Và B=(-1/2).Hãy so sánh Avà B
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SSH:(20152-12):10+1=2015
(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152
-10+(-10)+(-10)+...+(-10)+20152
-10x(2015-1):2+20152=12
=> C=12
\(1,\\ a,=\left[x^3\left(x-2\right)-4x\left(x-2\right)\right]:\left(x^2-4\right)\\ =x\left(x^2-4\right)\left(x-2\right):\left(x^2-4\right)=x\left(x-2\right)\\ b,=\left(2014-14\right)^2=2000^2=4000000\\ 2,\\ A=2015\cdot2013\cdot\left(2014^2+1\right)\\ A=\left(2014^2-1\right)\left(2014^2+1\right)\\ A=2014^4-1< B=2014^4\)
https://olm.vn/cau-hoi/a-cho-a12211216211002-ctr-a12-b-cho-p122132142120232-ctr-p-khong-la-so-tu-nhien-c-cho-c132152172120211.8293222842881
Cô làm rồi em nhá
Câu a, xem lại đề bài
Câu b:
P = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ...+ \(\dfrac{1}{2023^2}\)
Vì \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)
........................
\(\dfrac{1}{2023^2}\) < \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)
Cộng vế với vế ta có:
0< P < 1 - \(\dfrac{1}{2023}\) < 1
Vậy 0 < P < 1 nên P không phải là số tự nhiên vì không tồn tại số tự nhiên giữa hai số tự nhiên liên tiếp
Câu c:
C = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{2021^2}\) + \(\dfrac{1}{2023^2}\) = C
B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+ \(\dfrac{1}{2020^2}\) + \(\dfrac{1}{2023^2}\) > 0
Cộng vế với vế ta có:
C+B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\)+ \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{2023^2}\) > C + 0 = C > 0
Mặt khác ta có:
1 > \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\) (cm ở ý b)
Vậy 1 > C > 0 hay C không phải là số tự nhiên (đpcm)
Đặt S=1/6+1/12+1/20+1/30+1/42+1/56+1/72
=> S=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> S=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
=> S=1/2-1/9
=> S=7/18
Vì 7/18<1/2
=> S<1/2
Mọi người k mik nhé, :)))
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/8-1/9
= 1/2 - 1/9
= 7/18
Bn tự so sánh vs 1/2 nha
Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)
\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)
\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)
Mà \(\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)
Vậy : M > N
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(\Leftrightarrow A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{4052169}-1\right)\left(\frac{1}{\text{}\text{}4056196}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-4056195}{\text{}4056196}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3}{2^2}.\frac{\left(-2\right)4}{3^3}.\frac{\left(-3\right)5}{4^2}.....\frac{\left(-2013\right)2015}{\text{}2014^2}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(-2\right)....\left(-2013\right)}{2.3...1014}.\frac{3.4......2015}{2.3......2014}\)
\(\Leftrightarrow A=\frac{-1}{1014}.\frac{2015}{2}=\frac{-2015}{4028}\)
VÌ \(\frac{-2015}{4028}< \frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\Leftrightarrow A< B\)
Ta có \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-4056195}{2014^2}\)
\(=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{2013.2015}{2014^2}\right)=-\left(\frac{1.3.2.4...2013.2015}{2.2.3.3...2014.2014}\right)\)
\(=-\left(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\right)=-\frac{2015}{2014.2}=-\frac{2015}{4028}< \frac{-2014}{4028}< \frac{1}{2}=B\)
=> A < B