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ta có \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).......\left(\frac{1}{10}-1\right)\)
\(A=-\left(\frac{1}{2}.\frac{2}{3}.....\frac{9}{10}\right)\)
\(A=-\frac{1}{10}\)
vi\(-\frac{1}{10}>-\frac{1}{9}\)
do đó A>\(\frac{-1}{9}\)
Bài này dễ mà bạn cũng hỏi =(((
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{400}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-399}{400}\)
\(=\frac{1.\left(-3\right)}{2.2}.\frac{2.\left(-4\right)}{3.3}.\frac{3.\left(-5\right)}{4.4}....\frac{19.\left(-21\right)}{20.20}\)
\(=\frac{\left(1.2.3...19\right).\left(\left(-3\right).\left(-4\right).\left(-5\right)...\left(-21\right)\right)}{\left(2.3.4...20\right)\left(2.3.4...20\right)}=\frac{1}{20}.\frac{\left(-21\right)}{2}=\frac{-21}{40}\)
Dễ dàng nhận thấy \(\frac{21}{40}>\frac{1}{2}\Rightarrow\frac{-21}{40}< \frac{-1}{2}\)
Vậy \(A< -\frac{1}{2}\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{400}-1\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{399}{400}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{19.21}{20.20}\)
\(=\frac{\left(1.2.3...19\right)\left(3.4.5...21\right)}{\left(2.3.4....20\right)\left(2.3.4....20\right)}\)
\(=\frac{1.21}{20.2}=\frac{21}{40}\)
Dễ thấy \(\frac{21}{40}>\frac{-1}{2}\)
Vậy A > -1/2
Nhầm rồi :v, làm lại
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{400}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot\cdot\cdot\cdot\frac{-399}{400}\)
\(=\frac{1.\left(-3\right)}{2.2}\cdot\frac{2.\left(-4\right)}{3.3}\cdot\cdot\cdot\cdot\frac{19.\left(-21\right)}{20.20}\)
\(=\frac{\left(1.2....19\right).\left[-\left(3.4.5...21\right)\right]}{\left(2.3....20\right)\left(2.3....20\right)}\)
\(=\frac{1.\left(-21\right)}{20.2}=\frac{-21}{40}\)
Dễ thấy \(\frac{21}{40}>\frac{20}{40}\Rightarrow\frac{-21}{40}< \frac{-20}{40}=\frac{-1}{2}\)
Vậy A < -1/2
A = 1/1×2 + 1/2×3 + 1/3×4 + .. + 1/99×100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
A = 1 - 1/100 < 1
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}< 1\)
=> ĐPCM
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)...\left(\frac{19}{19}-\frac{1}{19}\right)\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1.2.3...18.19}{2.3.4...19.20}\)
\(A=\frac{1}{20}\Leftrightarrow A>\frac{1}{21}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}......\frac{19}{20}=\frac{1}{20}>\frac{1}{21}\)
\(\text{Vậy: A lớn hơn 1/21}\)
xét (1/4-1)*(1/9-1)*(1/16-1)*...*(1/400-1)
= \(-\frac{3}{4}\times\frac{-8}{9}\times-\frac{15}{16}\times.....\times-\frac{399}{400}\)
=\(-\frac{3}{2^2}\times\left(\frac{-8}{3^2}\right)\times\left(\frac{-15}{4^2}\right)\times........\times\left(\frac{-399}{20^2}\right)\)
dãy trên có số số hạng là:( 20-2):1+1=19(số hạng)
mà các số đều là các số âm => có 19 số âm nhân vào nhau sẽ ra số âm
Vậy A< 1/2
tk mình nha bạn cũ