1) So sánh A và B
A=6100.(\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+...+\(\frac{1}{3^{99}}\))+4.299
B=299.(3100+1)
a,A=\(2\frac{1}{2}:\left(\frac{-1}{2}\right)^2-\frac{1}{-3}.\left(\frac{-1}{2}-\frac{4}{3}:\frac{-8}{9}\right)\)
b,B=\(\left(3\frac{10}{99}+4\frac{11}{99}-\frac{58}{299}\right).\left(\frac{1}{2}-\frac{4}{3}-\frac{1}{6}\right)\)
Trả lời
B=(3 10/99+4 11/99-58/299).(1/2-4/3-1/6)
=(......................................).0
=0.
1. CMR:
C = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}< \frac{1}{2}\)\(\frac{1}{2}\)
2. So sánh
a) 9920 và 99910
b) 920 và 2713
so sánh :
A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.......+\frac{1}{3^{99}}\) với B = \(\frac{1}{2}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
Ta đặt \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Ta so sánh giữa A và C.
\(\frac{1}{3}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{3^3}< \frac{1}{3.4};....;\frac{1}{3^{99}}< \frac{1}{99.100}\Leftrightarrow A< C\)( 1 )
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Mà \(\frac{99}{100}< \frac{1}{2}\Rightarrow C< B\)( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow A< C< B\Leftrightarrow A< B\)
A=1/3 +1/3^2+...+1/3^99
-> 3A= 1+1/3+1/3^2+...+1/3^98
-> 3A-A=2A=1-1/3^99
A=(1-1/3^99)/2. vì 1-1/3^99 < 1 -> A< B. Vậy A < B
\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{99}}+\frac{1}{3^{100}}\) và \(\frac{1}{2}\)
So sánh
ta có: \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}+\frac{1}{3^{101}}\)
\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^{101}}< \frac{1}{3}\)
\(\Rightarrow\frac{2}{3}A< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)
\(\Rightarrow A< \frac{1}{2}\)
so sánh \(\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.............+\frac{1}{3^{99}}\right)và\frac{1}{2}\)
So sánh:
a) \(\frac{-3}{1.3}+\frac{-3}{3.5}+...+\frac{-3}{97.99}\)và \(\frac{49}{-20}\)
b)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}và\frac{99}{202}\)
c)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}và\frac{99}{100}\)
a,\(\frac{-3}{1.3}+\frac{-3}{3.5}+....+\frac{-3}{97.99}\)
= -3.\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
=\(\frac{-3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{-3}{2}\left(1-\frac{1}{99}\right)\)
=\(\frac{-3}{2}.\frac{98}{99}\)
=\(\frac{49}{-33}\)>\(\frac{49}{-20}\)
so sánh
\(\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)và\frac{1}{2}\)
Đặt \(K=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)
\(3K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)
\(3K-K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\)
\(2K=\)\(1-\frac{1}{3^{99}}\)
\(K=\frac{1-\frac{1}{3^{99}}}{2}\)
Có \(1-\frac{1}{3^{99}}\) < \(\frac{1}{2}\)
\(\Rightarrow K\) < \(\frac{1}{2}\)
Vậy \(\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\) < \(\frac{1}{2}\)
So sánh A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{98^2}+\frac{1}{99^2}\) và B=\(\frac{304}{1975}\)
Bài toán : So sánh A với \(\frac{1}{3}\)
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
#)Giải :
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)
\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)
\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)
Có lẽ đúng .........................
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
\(\Rightarrow A+\frac{1}{3}A=\frac{1}{3^1}+\left(\frac{-1}{3^{101}}\right)=\frac{1}{3^1}-\frac{1}{3^{101}}\)
\(\Rightarrow A\left(1+\frac{1}{3}\right)=\frac{1}{3^1}-\frac{1}{3^{101}}\)
\(\Rightarrow\frac{4}{3}A=\frac{1}{3^1}-\frac{1}{3^{101}}\)
\(A=\left(\frac{1}{3^1}-\frac{1}{3^{101}}\right):\frac{4}{3}\)
\(A=\left(\frac{1}{3^1}-\frac{1}{3^{101}}\right).\frac{3}{4}\)
\(A=\frac{1}{3^1}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)
\(A=\frac{1}{4}-\frac{1}{3^{100}.4}< \frac{1}{4}< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{3}\)
Vậy \(A< \frac{1}{3}\)