Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{240}\right)\)
\(A=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+....+\dfrac{1}{15\times16}\right)\)
\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)
b) cậu đi tìm số sốm hạng là : \(\left(2010-1\right):1+1=2010\)
\(\Rightarrow\)số cặp trong phép tính là : \(2010:2=1005\)(cặp)
\(\Rightarrow B=1-2+3-4+...+2009-2010\)(1005 cặp)
\(\Rightarrow\left(1-2\right)+\left(3-4\right)+...+\left(2009-2010\right)\)
\(\Rightarrow B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)(1005 số -1)
\(\Rightarrow B=\left(-1\right).1005\)
\(\Rightarrow B=\left(-1005\right)\)
cậu tik cho mik nhé!!!
Ta có:
\(2007A=\dfrac{2007^{2009}+2007}{2007^{2009}+1}=1+\dfrac{2006}{2007^{2009}+1}\)\(2007B=\dfrac{2007^{2010}+10}{2007^{2010}+1}=1+\dfrac{9}{2007^{2010}+1}\)Vì \(\dfrac{2007}{2007^{2009}+1}>\dfrac{2007}{2007^{2010}+1}\)
=>2007A > 2007B
Do đó A>B
Vậy A>B
Ta có : \(B\) = \(\dfrac{2007^{2009}+1}{2007^{2010}+1}\) \(< 1\) \(\Rightarrow\dfrac{2007^{2009}+1}{2007^{2010}+1}< \dfrac{2007^{2009}+1+2006}{2007^{2010}+1+2006}\) \(=\dfrac{2007^{2009}+2007}{2007^{2010}+2007}\)
\(=\dfrac{2007\left(2007^{2008}+1\right)}{2007\left(2007^{2009}+1\right)}\) \(=\dfrac{2007^{2008}+1}{2007^{2009}+1}=A\)
Vậy \(A>B\)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì :
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
hay A > B
Vậy A > B
1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
a) Xét:
\(a>b\)
\(\Rightarrow\dfrac{a}{b}>1\Rightarrow\dfrac{a+m}{b+m}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{a+m}\)
\(a< b\)
\(\Rightarrow\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
\(a=b\)
\(\Rightarrow\dfrac{a}{b}=1\Rightarrow\dfrac{a+m}{b+m}=1\Rightarrow\dfrac{a}{b}=\dfrac{a+m}{b+m}=1\)
Mk chỉ áp dụng tính 1 câu,câu sau làm tương tự
b)
Ta có:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{1993}+1}{10^{1992}+1}< 1\)
\(B< \dfrac{10^{1993}+1+9}{10^{1992}+1+9}\Rightarrow B< \dfrac{10^{1993}+10}{10^{1992}+10}\Rightarrow B< \dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\Rightarrow B< \dfrac{10^{1992}+1}{10^{1991}+1}=A\)
\(B< A\)
@@ ~ học tốt ~