So Sánh A vs B :
A=\(\frac{100^{2009}+1}{100^{2008}+1}\) B=\(\frac{100^{2010}+1}{100^{2009}+1}\)
A=\(\frac{5^0+5^1+5^2+5^3+.....+5^9}{5^0+5^1+5^2+.....+5^8}\) B=\(\frac{3^0+3^1+3^2+.....+3^9}{3^0+3^1+3^2+.....+3^8}\)
Những câu hỏi liên quan
a) So sánh: A=\(\frac{100^{2009}+1}{100^{2008}+1}\)và B=\(\frac{100^{2010}+1}{100^{2009}+1}\)
b) Chứng minh rằng: \(\frac{1}{6}\)<\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{4}\)
\(a)\) 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 ~
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1.Tính tổngSleft(frac{-1}{7}right)^0+left(frac{-1}{7}right)^1+left(frac{-1}{7}right)^2+...+left(frac{-1}{7}right)^{2007}2.Tìm x5^x+5^{x+2}6503.CMRfrac{1}{6} frac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2} frac{1}{4}4. Cho Afrac{1}{2010}+frac{2}{2009}+frac{3}{2008}+...+frac{2009}{2}+frac{2010}{1} Bfrac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2010}+frac{1}{2011}So sánh A và B
Đọc tiếp
1.Tính tổng
\(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
2.Tìm x
\(5^x+5^{x+2}=650\)
3.CMR
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
4. Cho \(A=\frac{1}{2010}+\frac{2}{2009}+\frac{3}{2008}+...+\frac{2009}{2}+\frac{2010}{1}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2010}+\frac{1}{2011}\)
So sánh A và B
a,A=1-2+3+4-5-6+7+8-9-...+2007+2008-2009-2010
b, \(\frac{1}{5^2}-\frac{1}{5^3}+\frac{1}{5^4}-\frac{1}{5^5}+..-\frac{1}{5^{101}}\).CM<\(\frac{1}{30}\)
cho biểu thức :
A=\(\frac{1+9^2+9^3+...+9^{2010}}{1+9+9^2+...+9^{2009}}\)
B=\(\frac{1+5^1+...+5^{2010}}{1+5+5^2+...+5^{2009}}\)
so sánh A vàB
A = \(1+\frac{9^{2010}}{1+9+9^2+....+9^{2009}}\)= \(1+1:\frac{1+9+9^2+....+9^{2009}}{9^{2010}}\)= \(1+1:\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+\frac{1}{9^{2008}}+...+\frac{1}{9}\right)\)
B = \(1+\frac{5^{2010}}{1+5+5^2+....+5^{2009}}\)= \(1+1:\frac{1+5+5^2+...+5^{2009}}{5^{2010}}\)= \(1+1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)
Do \(\frac{1}{9^{2010}}
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Cho biểu thức: A =\(\frac{1+9+9^2+...+9^{2010}}{1+9+9^2+...+9^{2009}}\) B =\(\frac{1+5+5^2+...+5^{2010}}{1+5+5^2+...+2^{2009}}\)
Hãy so sánh A và B ???
Đặt M = \(1+9+9^2+......+9^{2010}\)
\(9M=9+9^2+9^3+......+9^{2011}\)
\(9M-M=8M=9^{2011}-1\)
Đặt K = \(1+9+9^2+......+9^{2009}\)
\(9K=9+9^2+9^3+.....+9^{2010}\)
\(9K-K=8K=9^{2010}-1\)
\(\Rightarrow A=\frac{9^{2011}-1}{9^{2010}-1}\)
Đặt H=\(1+5+5^2+....+5^{2010}\)
\(5H=5+5^2+......+5^{2011}\)
\(5H-H=4H=5^{2011}-1\)
ĐẶT G = \(1+5+5^2+.......+5^{2009}\)
\(5G-G=4G=5^{2010}-1\)
\(\Rightarrow B=\frac{5^{2011}-1}{5^{2010}-1}\)
Rồi bạn so sánh sẽ ra ngay
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So sánh
a)A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)và B=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
b)M=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)và N=\(\frac{2009^{2009}-2}{2009^{2010}-2}\)
c)P=\(\frac{1+5+5^2+5^3+...+5^{10}}{1+5+5^2+5^3+...+5^9}\)và Q=\(\frac{1+3+3^2+3^3+...+3^{10}}{1+3+3^2+3^3+...+3^9}\)
a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)
\(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B Vay A<B
b,lam tuong tu nhu y a
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so sánh mà ko tính kết quả cụ thể:
a=2009^2008+1 phần 2009^2009+1
b=2009^2009+1 phần 2009^2010+1
b/ c=1*3*5*7...*99
d=51 phần 2 *52 phần 2* 53 phần 2*....*100 phần 2
Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
Xem thêm câu trả lời
Bài1:
1-2+3+4-5-6+7+8-9-.......+2007+2008-2009-2010
Bài 2:
Cho a,b thuộc N*. CMR: a,b là 2 số NTCN thì 7a+5b và 4a+3b cũng là hai số NTCN
Bài 3: Cho S= 1/(5^2) - 2/(5^3) + 3/(5^4) - 4/(5^5) + 99/(5^100) - 100/(5^101)
CMR: S< 1/36
