Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:\(A< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Mặt khác:\(A>\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
Vậy \(\frac{8}{9}>A>\frac{2}{5}\)
A = 1 / 2.2 + 1 / 3.3 + 1 / 4.4 + .... + 1 / 9.9
A < 1/1.2 + 1/2.3 + .....+ 1/8.9
A < 1 - 1/2 + 1/2 - 1/3 + ......+ 1/8 - 1/9
A < 1 - 1/9
=> A < 8/9 (1)
Mặt khác ta có:
A > 1/2.3 + 1/3.4 +.....+ 1/9.10
A > 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/9 - 1/10
A > 1/2 - 1/10
A > 4/10
=> A > 2/5 (2)
Từ (1) và (2) => 8/9 > A > 2/5
**** K mk nha các bn! đúng 100000% lun đó!!!!!!!!!
\(B=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(B=\frac{1.2+2^2.1.2+3^21.2+4^2.1.2+5^2.1.2}{3.4+2^23.4+3^23.4+4^23.4+5^23.4}\)
\(B=\frac{2.\left(1+2^2+3^2+4^2+5^2\right)}{12\left(1+2^2+3^2+4^2+5^2\right)}\)\(\Rightarrow B=\frac{2}{12}=\frac{1}{6}\)
\(A=1+\frac{5^9}{1+5+..+5^8}\)
\(=1+\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}\)
Tương tự:
\(B=1+\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
Vì \(\frac{1}{5}< \frac{1}{3}\) , \(\frac{1}{5^2}< \frac{1}{3^2}\), . . .
nên: \(\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}>\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
=> A > B
Vậy đề bạn cho chứng minh A < B là sai nhé.
Ta có:\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
=>\(A=\frac{\left(1+5+5^2+...+5^8\right)}{\left(1+5+5^2+...+5^8\right)}+\frac{5^9}{1+5+5^2+...+5^8}\)
=>\(A=1+\frac{5^9}{1+5+5^2+...+5^8}\)
Ta có:\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
=>\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}\)
=>\(B=1+\frac{3^9}{1+3+3^2+...+3^8}\)
vì:\(1+3+3^2+...+3^8< 1+5+5^2+...+5^8\)
Nên A<B(đpcm).
A=1/1+5+5^2+5^3+...+5^8+5+5^2+5^3+...+5^9=1/1+5+5^2+5^3+...+5^8+5.
Tương tự B=1/1+3+3^2+...+3^8+3
=>A>B.
k nha.
\(\frac{1}{2^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
cm tt => đpcm
\(\frac{1}{2^2}<\frac{1}{1.2}=1-\frac{1}{2}\)
cmtt =>...................
ta có A=1/2^2+1/3^2+1/4^2+...+1/9^2
mà 1/2^2>1/2.3=1/2-1/3
1/3^2>1/3.4=1/3-1/4
1/4^2>1/4.5=1/4-1/5
........
1/9^2>1/9.10=1/9-1/10
=> 1/2^2+1/3^2+1/4^2+...+1/9^2>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
=>1/2^2+1/3^2+1/4^2+...+1/9^2>1/2-1/10=2/5
vậy A>2/5 *
ta có 1/2^2<1/1.2=1-1/2
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
.......
1/9^2<1/8.9=1/8-1/9
=>1/2^2+1/3^2+1/4^2+...+1/9^2<1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
=>1/2^2+1/3^2+1/4^2+...+1/9^2<1-1/9=8/9
vậy A<8/9 **
từ *,** => 8/9>A>2/5 (đpcm)
Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+.................+\dfrac{1}{9^2}\)
Xét :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{2^3}< \dfrac{1}{2.3}\)
..................................
\(\dfrac{1}{9^2}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...............+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}=\dfrac{8}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\rightarrowđpcm\) \(\left(1\right)\)
Xét :
\(\dfrac{1}{2^2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{2^3}>\dfrac{1}{3.4}\)
......................
\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+.............+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\rightarrowđpcm\)\(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\rightarrowđpcm\)
~ Chúc bn học tốt ~