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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1}{20}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{18}{19}.\frac{19}{20}\)
\(\Leftrightarrow A=\frac{1}{20}>\frac{1}{21}\)
\(\Leftrightarrow A>\frac{1}{21}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)................\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow B=\frac{3}{4}.\frac{8}{9}..................\frac{99}{100}\)
\(B=\frac{1.3}{2^2}.\frac{2.4}{3^2}................\frac{9.11}{50^2}\)
\(B=\frac{11}{50}< \frac{11}{21}\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot...\left(\frac{1}{10}-1\right)\)
\(A=\left(\frac{1}{2}-\frac{2}{2}\right)\left(\frac{1}{3}-\frac{3}{3}\right)\cdot...\cdot\left(\frac{1}{10}-\frac{10}{10}\right)\)
\(A=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{9}{10}\right)\)
\(A=\frac{-1}{2}\cdot\frac{-2}{3}\cdot...\cdot\frac{-9}{10}\)
\(A=\frac{\left(-1\right)\cdot\left(-2\right)\cdot...\cdot\left(-9\right)}{2\cdot3\cdot...\cdot10}\)
\(A=\frac{\left(-1\right)\cdot2\cdot...\cdot9}{2\cdot3\cdot...\cdot10}=\frac{-1}{10}\)
Mà \(\frac{-1}{10}>\frac{-1}{9}\)nên A > -1/9
Phần cuối tương tự
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{100}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...............\frac{99}{100}\)
\(=\frac{3.8.15......99}{4.9.16....100}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).......\left(9.11\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)......\left(10.10\right)}\)
\(=\frac{\left(1.2.3.....9\right).\left(3.4.5......11\right)}{\left(2.3.4.....10\right).\left(2.3.4.......10\right)}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{21}\)
Vậy B<11/21
\(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}=-\frac{3.8...99}{4.9....100}\)
\(=-\frac{1.3.2.4...9.11}{2.2.3.3....10.10}=-\frac{\left(1.2...9\right).\left(3.4...11\right)}{\left(2.3...10\right).\left(2.3...10\right)}=-\frac{1.11}{10.2}=-\frac{11}{20}< -\frac{11}{21}\)
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\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)
\(B=\frac{1}{10}.\frac{11}{2}\)
\(B=\frac{11}{20}>\frac{11}{21}\)
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
a, Ta có :
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...........\left(\dfrac{1}{10}-1\right)\)
\(=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right).........\left(\dfrac{1}{10}-\dfrac{10}{10}\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}...............\dfrac{-9}{10}\)
\(=\dfrac{-1.\left(-2\right)............\left(-9\right)}{2.3........9.10}\)
\(=\dfrac{-1}{10}< \dfrac{-1}{9}\)
\(\Leftrightarrow A< \dfrac{-1}{9}\)
b, \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)..........\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right).........\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-99}{100}\)
\(=\dfrac{1.\left(-3\right).2\left(-4\right)............9\left(-11\right)}{2^2.3^2.......10^2}\)
\(=\dfrac{1.2.3........9}{2.3.......10}.\dfrac{\left(-3\right)\left(-4\right)....\left(-11\right)}{2.3...10}\)
\(=\dfrac{1}{10}.\dfrac{-11}{1}\)
\(=\dfrac{-11}{10}>\dfrac{-11}{21}\)
\(\Leftrightarrow B>\dfrac{-11}{21}\)
Ta có:
$$B = \left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\dots\left(\frac{1}{100}-1\right)$$$$B = \left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\dots\left(\frac{-99}{100}\right)$$$$B = -\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2 \cdot 4}{3 \cdot 3} \dots \frac{9 \cdot 11}{10 \cdot 10}$$$$B = -\frac{(1 \cdot 2 \dots 9) \cdot (3 \cdot 4 \dots 11)}{(2 \cdot 3 \dots 10) \cdot (2 \cdot 3 \dots 10)}$$$$B = -\frac{1 \cdot 11}{10 \cdot 2}$$$$B = -\frac{11}{20}$$Vì $-\frac{11}{20} < -\frac{11}{21}$ nên $B < -\frac{11}{21}$.
Kết luận: $B < -\frac{11}{21}$
no name
:/ jv
Ta có: \(B=\left(\frac14-1\right)\left(\frac19-1\right)\cdot\ldots\cdot\left(\frac{1}{100}-1\right)\)
\(=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{10^2}-1\right)\)
\(=\left(\frac12-1\right)\cdot\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{10}-1\right)\cdot\left(\frac12+1\right)\left(\frac13+1\right)\cdot\ldots\cdot\left(\frac{1}{10}+1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-9}{10}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{11}{10}=-\frac{1}{10}\cdot\frac{11}{2}=\frac{-11}{20}\)
20<21
=>\(\frac{11}{20}>\frac{11}{21}\)
=>\(-\frac{11}{20}<-\frac{11}{21}\)
=>\(B<-\frac{11}{21}\)
\(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\dots \left(\frac{1}{100}-1\right)\)Mỗi thừa số có dạng tổng quát là \(\left(\frac{1}{n^2} - 1\right)\) với \(n\) chạy từ \(2\) đến \(10\). Ta biến đổi:
\(\frac{1}{n^{2}}-1=\frac{1-n^{2}}{n^{2}}=-\frac{n^{2}-1}{n^{2}}=-\frac{(n-1)(n+1)}{n\cdot n}\)Số lượng các thừa số từ \(n=2\) đến \(n=10\) là \(10 - 2 + 1 = 9\) thừa số. Vì 9 là số lẻ, nên tích \(B\) sẽ có dấu âm.2. Tính giá trị \(B\)\(B=(-1)^{9}\cdot \left[\frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\dots \frac{9\cdot 11}{10\cdot 10}\right]\)Triệt tiêu các số hạng ở tử và mẫu:
- Các số từ \(2\) đến \(9\) ở tử số xuất hiện 2 lần, triệt tiêu hết với mẫu.
- Còn lại số \(1\) và \(11\) ở tử; số \(2\) và \(10\) ở mẫu.
\(B=-\frac{1\cdot 11}{2\cdot 10}=-\frac{11}{20}\)3. So sánh \(B\) với \(-\frac{11}{21}\)Ta so sánh hai phân số âm: \(-\frac{11}{20}\) và \(-\frac{11}{21}\).- Vì \(20 < 21\) nên \(\frac{11}{20} > \frac{11}{21}\).
- Khi nhân với \(-1\), dấu bất đẳng thức đổi chiều: \(-\frac{11}{20} < -\frac{11}{21}\).
Kết luận: \(B < -\frac{11}{21}\)B = (1/4 - 1)(1/9 - 1)...(1/100 - 1)
= (1/2² - 1)(1/3² - 1)...(1/10² - 1)
= (-1.3/2²)(-2.4/3²)...(-9.11/10²)
Có 9 thừa số âm nên:
B = -1.3.2.4...9.11/(2².3²...10²)
= -11/20
Ta có:
-11/20 < -11/21
Vậy B < -11/21, vì -11/20 là số âm có giá trị nhỏ hơn -11/21.