Cho S = 1/101 + 1/202 + 1/203 + ... + 1/299 + 1/300. Chứng minh rằng 2/3 < S < 2
Cho S = 1/201 + 1/202 + 1/203 + ... + 1/299 + 1/300. Chứng minh rằng 1/3 < S < 1/2
Cho S = 1/201 + 1/202 + 1/203 + ... + 1/299 + 1/300. Chứng minh rằng 1/3 < S < 1/2
1/201>1/300,1/202>1/300.................1/300=1/300 =>S>1/300.100=1/3 1/201<1/200, 1/202<1/300.................1/300<1/200=>S<1/200.100=1/2
s=1/201+1/202+1/203+.........+1/299+1/300 Chứng tỏ S>11/30
Chứng minh rằng:
1/101+1/102+...+1/299+1/300>2/3
S= 1/201 + 1/202 + ... + 1/ 299 + 1/300 chứng tỏ S > 11/3o
Cho A = 1/200+1/201+1/202+1/203+.......+1/300 . Chứng minh rằng A <9/20
Chứng tỏ rằng 1/101+1/102+....+1/299+1/300>2/3
chứng tỏ rằng 1/101+1/102+........+1/299+1/300>2/3
Tra lời:
Ta có:
1/101➢1/300+1/102➢1/300+1/103➢1/300+1/104➢1/300+.....+1/299➢1/300
=1/101+1/102+1/103+...1/299➢199/300
=1/101+1/102+1/103+...1/299+1/300➢199/300+1/300
=200/300=2/3.
Note: ➢ là dau lớn do nhe. Nho tick cho minh nha😊😉
chứng tỏ rằng 1/101+1/102+...+1/299+1/300>2/3
\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\). . . . \(+\)\(\frac{1}{299}\)\(+\)\(\frac{1}{300}\)\(\ge\)\(\frac{2}{3}\)\(\ge\)\(\frac{1}{300}\)\(+\)\(\frac{1}{300}\)\(+\)\(\frac{1}{300}\)\(=\)\(\frac{200}{300}\)\(=\)\(\frac{2}{3}\)
do \(\frac{1}{101}\)..... \(\frac{1}{300}\)có 200 số
\(\Rightarrow\)\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\)..... \(+\)\(\frac{1}{299}\)\(+\)\(\frac{1}{300}\)\(\ge\)\(\frac{1}{300}\)\(\times\)200
\(\ge\)\(\frac{2}{3}\)