Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dễ thấy A < 1. Áp dụng nếu \(\frac{a}{b}<1\) thì \(\frac{a}{b}<\frac{a+m}{b+m}\) ta có :
\(A=\frac{100^{100}+1}{100^{99}+1}<\frac{\left(100^{100}+1\right)+\left(100^{31}-1\right)}{\left(100^{99}+1\right)+\left(100^{31}-1\right)}=\frac{100^{100}+100^{31}}{100^{99}+100^{31}}=\frac{100^{31}.\left(100^{69}+1\right)}{100^{31}.\left(100^{68}+1\right)}=\frac{100^{69}+1}{100^{68}+1}=B\)
Vậy A < B
\(\frac{100^{100}+1}{100^{99}+1}=\frac{100^{69}+1}{100^{68}+1}\)
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
A=1+4+42+43+...+499
=>4A=4+42+43+44+...+4100
=>4A-A=(4+42+43+44+...+4100)-(1+4+42+43+...+499)
=>3A=4100-1
=>A=\(\frac{4^{100}-1}{3}\) < 4100
=>A<B
\(A=1+4+4^2+4^3+...+4^{99}\)
=> \(4A=4+4^2+4^3+4^4+...+4^{100}\)
=> \(4A-A=\left(4+4^2+4^3+...+4^{100}\right)-\left(1+4+4^2+...+4^{99}\right)\)
=> \(3A=4^{100}-1\)
=> \(A=\frac{4^{100}-1}{3}\)
Ta có : \(B=4^{100}\) => \(\frac{B}{3}=\frac{4^{100}}{3}\)
Vì \(4^{100}-1<4^{100}\) => \(\frac{4^{100}-1}{3}<\frac{4^{100}}{3}\) => \(A<\frac{B}{3}\) (đpcm)
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(=>2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(=>2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(=>A=1-\frac{1}{2^{100}}\)
Ta có : \(1>\frac{1}{2^{100}}=>A>1-1=0\)
\(\frac{1}{2^{100}}>0=>1-\frac{1}{2^{100}}< 1-0=1\)
\(=>0< A< 1\)
Chúc bạn học tốt!
Dễ thấy A>0(vì 1/2>0;1/2^2>0;...;1/2^100>0 =>1/2+1/2^2+1/2^3+...+1/2^100>0)
2A=1+2/2^2+2/2^3+...+2/2^100(rút gọn 1 bước)
2A=1+1/2+1/2^2+...+1/2^99
2A-A=(1+1/2+1/2^2+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^99+1/2^100)
A=1-1/2^100<1
Vậy A<1
Cậu tự KL nhé
\(x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.\frac{9}{10}=\frac{63}{256}< \frac{63}{210}=0,3\)
\(x=\sqrt{0,1}>\sqrt{0,09}=0,3\)
=> y<x
Giải:
\(\frac{13}{20}=\frac{13.101}{20.101}=\frac{1313}{2020}\)
\(\frac{100}{101}=\frac{100.20}{101.20}=\frac{2000}{2020}\)
Vì \(1313<2000\Rightarrow\frac{1313}{2020}=\frac{2000}{2020}\Rightarrow\frac{13}{20}<\frac{100}{101}\)
Chúc bạn học tốt!
Bài 3:
\(\left(\dfrac{1}{32}\right)^7=\dfrac{1^7}{32^7}=\dfrac{1}{32^7}=\dfrac{1}{\left(2^5\right)^7}=\dfrac{1}{2^{35}}\\ \left(\dfrac{1}{16}\right)^9=\dfrac{1^9}{16^9}=\dfrac{1}{16^9}=\dfrac{1}{\left(2^4\right)^9}=\dfrac{1}{2^{36}}\)
Vì \(2^{35}< 2^{36}\) nên \(\dfrac{1}{2^{35}}>\dfrac{1}{2^{36}}\) hay \(\left(\dfrac{1}{32}\right)^7>\left(\dfrac{1}{16}\right)^9\)
A=-1/2*-2/3*-3/4*..*-2013/2014
A=-1*-2*-3*...*-2013/2*3*4*...*2014
A=-1/2014
ta có(-1)^2015=-1
B=-1/2015>-1/2014=A
nên A<B
Áp dụng nếu \(\frac{a}{b}>1\) thì \(\frac{a}{b}>\frac{a+m}{b+m}\) (m \(\in\) N*) ta có :
\(A=\frac{100^{1000}}{100^{900}}>\frac{100^{1000}+1}{100^{900}+1}=B\)
Vậy A > B
Vì A>1 suy ra A+m<A suy ra B<A