So sánh:
\(A=\dfrac{1}{2x2}\)+\(\dfrac{1}{3x3}\)+\(\dfrac{1}{4x4}\)+......+\(\dfrac{1}{99x99}\)+\(\dfrac{1}{100x100}\)với B= \(\dfrac{99}{100}\)
Chứng minh:
C=\(\dfrac{1}{2x2}\)+\(\dfrac{1}{3x3}\)+\(\dfrac{1}{4x4}\)+.....+\(\dfrac{1}{100x100}\)<1
\(C=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{100\times100}\\ C< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ C< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ C< 1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
Cho A = \(\dfrac{1}{2}x\dfrac{3}{4}x\dfrac{5}{6}x...x\dfrac{99}{100};B=\dfrac{1}{10}\) So sánh: A và B.
Tham khảo:
https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10
a) rút gọn: \(\dfrac{4^5x9^4-2x6^9}{2^{10}x3^8+6^8x20}\)
b) Cho A=\(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\).So sánh A với 2
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
B=\(\dfrac{1}{11}\)+\(\dfrac{1}{11^2}\)+\(\dfrac{1}{11^3}\)+...+\(\dfrac{1}{11^{99}}\)+\(\dfrac{1}{11^{100}}\)
so sanh B với \(\dfrac{1}{10}\)
\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
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 A VÀ B
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{4}-...-2\cdot\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{50}\)
\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Rightarrow A=B\)
tớ giải chi tiết hơn nhá:
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=(\(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
A=\(\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
A=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
Vậy A=B
So sánh bt: \(M=\dfrac{100^{100}+1}{100^{99}+1};N=\dfrac{100^{101}+1}{100^{100}+1}\)
Ta có:
\(M=\dfrac{100^{100}+1}{100^{99}+1}\)
\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}\)
\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}\)
\(\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}\)
\(N=\dfrac{100^{101}+1}{100^{100}+1}\)
\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}\)
\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}\)
\(\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}\)
Mà: \(100^{101}>100^{100}\)
\(\Rightarrow100^{101}+100>100^{100}+100\)
\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)
\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)
\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)
\(\Rightarrow N< M\)