a=\(\sqrt{1+^{ }2011}^2+^{^{^{\frac{2011^2}{2012^2}}}+\frac{2011}{2012}}\)
tính A biết: \(a+\frac{1}{a+\frac{1}{a+\frac{1}{a+\frac{1}{a}}}}\)
Những câu hỏi liên quan
\(S=\sqrt{1+2010^2+\frac{2010^2}{2011^2}}+\frac{2010}{2011}+\sqrt{1+2011^2+\frac{2011^2}{2012^2}}+\frac{2011}{2012}+\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}\)
bài 1 :a) Tính M:\(\frac{\frac{7}{2012}+\frac{7}{9}-\frac{1}{4}}{\frac{5}{9}-\frac{3}{2012}-\frac{1}{2}}\)
b) So sánh A và B biết A =\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\);;; B =\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{17}\)
Tìm x biết:
a) \(^{2^x+2^{x+1}+2^{x+2}+2^{x+3}=480}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right).x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{2}{2011}+\frac{1}{2012}\)
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
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Chính Xác 100% là X=5
k cho mink nhé các pạn
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tính M
M = \(\frac{\frac{7}{2012}+\frac{7}{9}-\frac{1}{4}}{\frac{5}{9}-\frac{3}{2012}-\frac{1}{2}}\)
so sánh A và B biết
A = \(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\)
B = \(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}\)
Tính:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+..+\frac{1}{2011}}\)
So sánh
a)
\(A=\frac{2011^{2012}+4}{2011^{2012}-1}\)với \(B=\frac{2011^{2012}+1}{2011^{2012}-4}\)
b)
\(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}\)với \(\frac{1}{2}\)
Tìm x biết:
a) \(^{2^x+2^{x+1}+2^{x+2}+2^{x+3}=480}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right).x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)
\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
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b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)
\(\Rightarrow x=2013\)
Vậy \(x=2013\)
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\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{2011}+\frac{1}{2012}\)
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\cdot\cdot\cdot\frac{2}{2010}+\frac{1}{2011}\)
Tính \(\frac{B}{A}\)
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
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từ dề bài ta có :
\(\frac{B}{A}=\frac{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=2012\)
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Giải phương trình sau ( giải từng bước nhá các bạn )
a) ( 1 + 1/2 + 1/3 + ..... + 1/2011 + 1/2012 ) . 503x = 1 + 2014/2 + 2015/3 + ...... + 4023/2011 + 4024/2012
c) ( \(\frac{2011}{1\cdot11}\) + \(\frac{2012}{2\cdot12}+.....+\frac{2011}{100\cdot110}\) ) x = \(\frac{2011}{1\cdot110}+\frac{2011}{2\cdot102}+....+\frac{2011}{10\cdot110}\)
từng bước bao gồm cả lập luân luôn
a)\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\) (1)
\(A=\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\) (có 2011 số hạng)
nếu ta trừ một vào từng số hạng được tử số giống nhau
\(A-2011=\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{4024}{2012}-1\right)\)
\(A-2011=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}=2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(A-2011+2012=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)công 2012 hai vế
\(A+1=VP=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\left(1\right)\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\left(2\right)\)
Chia cả hai vế (2) cho: \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\Rightarrow503x=2012\)
\(x=\frac{2012}{503}\)
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