So sánh 2 phân số sau : \(\left(\dfrac{1}{32}\right)^7\)và \(\left(\dfrac{1}{16}\right)^9\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)....\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
So sánh B với \(\dfrac{11}{21}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
Cho biểu thức \(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
Hãy so sánh A với \(\dfrac{11}{19}\)
`A = 3/4 xx 8/9 xx ... xx 99/100`
`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`
`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`
`= 1/10 xx 11`
`= 11/10`.
Ta có: `11/10 > 1`
`11/19 < 1`.
`=> A > 11/19`.
so sánh
\(\left(\dfrac{1}{16}\right)^{200}\) và \(\left(\dfrac{1}{2}\right)^{1000}\)
\(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)
So Sánh : \(\left(\dfrac{1}{16}\right)^{200}\)và\(\left(\dfrac{1}{2}\right)^{1000}\)
16 = 24
(\(\dfrac{1}{16}\))200 = \(\dfrac{1}{2^{4.200}}\) = \(\dfrac{1}{2^{800}}\)= (\(\dfrac{1}{2}\))800
So sánh với (\(\dfrac{1}{2}\))1000
Hai phân số cùng tử số, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn
Suy ra: (\(\dfrac{1}{16}\))200 > (\(\dfrac{1}{2}\))1000
Ta có: \(\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2}\right)^{800}\)
mà \(\left(\dfrac{1}{2}\right)^{800}>\left(\dfrac{1}{2}\right)^{1000}\)
nên \(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)
A= \(\left(\dfrac{1}{2}-1\right)\)\(\left(\dfrac{1}{3}-1\right)\).........\(\left(\dfrac{1}{10}-1\right)\). So sánh A với \(\dfrac{-1}{9}\)
B= \(\left(\dfrac{1}{4}-1\right)\)\(\left(\dfrac{1}{9}-1\right)\)...........\(\left(\dfrac{1}{100}-1\right)\). So sánh B với \(\dfrac{-11}{21}\)
a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)
\(=-\dfrac{1}{10}\)
9<10
=>1/9>1/10
=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)
=>\(A>-\dfrac{1}{9}\)
b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)
\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)
20<21
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
So sánh : \(\left(\dfrac{1}{2}\right)^{12}\)và \(\left(\dfrac{1}{3}\right)^9\)
\(\left(\dfrac{1}{2}\right)^{12}=\left(\dfrac{1}{8}\right)^3\\ \left(\dfrac{1}{3}\right)^9=\left(\dfrac{1}{27}\right)^3\\ Ta\text{ }có:\dfrac{1}{8}>\dfrac{1}{27}\\ Vậy:\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)
\(\left(\dfrac{1}{2}\right)^{12}=\dfrac{1}{4096};\left(\dfrac{1}{3}\right)^9=\dfrac{1}{19683}\\ \Rightarrow\dfrac{1}{4096}>\dfrac{1}{19683}\\ \Rightarrow\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)
Số nguyên \(x\) thỏa mãn \(\left(\dfrac{3}{4}-\dfrac{2}{3}\right)+\dfrac{5}{6}< x\le\dfrac{4}{5}-\left(\dfrac{3}{10}-\dfrac{5}{4}\right)\) là:
A. \(x=1\) B. \(x=0\) C. \(x=2\) D. \(x\in\left\{0;1\right\}\)
So sánh 3 phân số: \(\dfrac{9}{170};\dfrac{9}{230};\dfrac{53}{144}\)
Câu 1: D
Câu 3: 53/144>9/170>9/230
So sánh \(\dfrac{3}{4}+\dfrac{3}{9}+\dfrac{3}{16}+...+\dfrac{3}{\left(3n\right)^2}\) với 1
A = \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\) + \(\dfrac{3}{25}\) +..............+ \(\dfrac{3}{(3n)^2}\)
A = ( \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\)+ \(\dfrac{3}{25}\)) +.....+ \(\dfrac{3}{(3n)^2}\)
A = 3. ( \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\))+............+ \(\dfrac{3}{(3n)^2}\)
A = 3.( \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + \(\dfrac{1}{5.5}\)) +............+ \(\dfrac{3}{(3n)^2}\)
Vì \(\dfrac{1}{2}\) > \(\dfrac{1}{3}\) > \(\dfrac{1}{4}\) > \(\dfrac{1}{5}\)Ta có : \(\dfrac{1}{2.2}>\dfrac{1}{2.3}>\dfrac{1}{3.3}>\dfrac{1}{3.4}>\dfrac{1}{4.4}>\dfrac{1}{4.5}>\dfrac{1}{5.5}>\dfrac{1}{5.6}\)
A > 3. ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)) + ............+ \(\dfrac{1}{(3n)^2}\)
A > 3. ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)) +.....+ \(\dfrac{1}{(3n)^2}\)
A > 3.( \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) +..............+ \(\dfrac{1}{(3n)^2}\)
A > 3. \(\dfrac{1}{3}\) +...............+ \(\dfrac{1}{(3n)^2}\)
A > 1 +..........+ \(\dfrac{1}{9n^2}\) > 1
A > 1
cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}-1\right).....\left(\dfrac{1}{10}-1\right)\). so sánh a với \(-\dfrac{1}{9}\)
Ta có:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)
\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)
\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)
\(A=-\dfrac{1}{10}\)
Mà: \(10>9\)
\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)
\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)
\(\Rightarrow A>-\dfrac{1}{9}\)