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a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.
Ta có:\(\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)...\left(1-\dfrac{2010}{2010}\right)\left(1-\dfrac{2011}{2010}\right)\)
=\(\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)....\)0\(\left(1-\dfrac{2011}{2010}\right)=0\)
=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)
\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
Đặt D1 = \(\dfrac{2010}{1}\) + \(\dfrac{2009}{2}\) + \(\dfrac{2008}{3}\) + ... + \(\dfrac{1}{2010}\)
= 1 + ( 1+ \(\dfrac{2009}{2}\)) + ( 1+ \(\dfrac{2008}{3}\)) + ... + (1+\(\dfrac{1}{2010}\))
= \(\dfrac{2011}{2}\) + \(\dfrac{2011}{3}\)+ ... + \(\dfrac{2011}{2010}\) + \(\dfrac{2011}{2011}\)
= 2011. ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\))
Đặt D2 = \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\)
=> D = 2011
cho mk 1 tick nha 
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
\(Q=\dfrac{1}{2011}+\dfrac{2}{2010}+\dfrac{3}{2009}+...+\dfrac{2010}{2}+\dfrac{2011}{1}\)
\(Q=\left(1+\dfrac{2}{2011}\right)\left(1+\dfrac{2}{2010}\right)+\left(1+\dfrac{3}{2009}\right)+...+\left(1+\dfrac{2010}{2}\right)+1\)
\(Q=\dfrac{2012}{2011}+\dfrac{2012}{2010}+\dfrac{2012}{2009}+...+\dfrac{2012}{2}+\dfrac{2012}{2012}\)
\(Q=2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(\Rightarrow\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)}=\dfrac{1}{2012}\)