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\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2007}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2008}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2008}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1004}\)
\(A=\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}\) (1)
\(B=\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}\) (2)
\(\left(1\right)\left(2\right)\Rightarrow\frac{A}{B}=\frac{\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}}{\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}}=1\)
a) 299A = \(1-\frac{1}{400}\) A= \(\frac{399}{400}\) :299
101B = \(1-\frac{1}{400}\) B = \(\frac{399}{400}\):101
\(\frac{A}{B}=\frac{299}{101}\)
Làm tắt ý a, mấy ý kia biết làm nhưng dài lắm
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 \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
Ta có: \(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Xét tử : \(2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+1+...+1\right)+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)( có 2008 số hạng 1 )
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)+1\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
Ghép tử và mẫu....
Vậy A = 2009
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 + 2^10]
Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]
Q = 2 . 3+2^3 .3 +... + 2^9 .3
Q = 3. [ 2 + 2^3 +... + 2^9]
Vậy Q chia hết cho 3
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