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A = 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/97.100
3A = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100 = (4-1)/1.4 + (7-4)/4.7 + (10-7)/7.10 + ... + (100-97)/97.100
= 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100 = 1 - 1/100 = 99/100
=> A = 33/100
A = x/2 => x = 2.A = 33/50
Ta có 1/1x4+1/4x7+...+1/2002x2005
<=> =1/3.3(1/1x4+1/4x7+...+1/2002x2005)
=1/3(3/1x4+3/4x7+...+3/2002x2005)
=1/3(1-1/4+1/4-1/7+...+1/2002-1/2005)
=1/3(1-1/2005)
=1/3.2004/2005
=1.2004/3.2005
=668/2005
\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+\(\frac{1}{2002.2005}\)=3(\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+ \(\frac{1}{2002.2005}\)):3=(\(\frac{3}{1.4}\)+ \(\frac{3}{4.7}\)+...+ \(\frac{3}{2002.2005}\)):3= (1-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)):3=(1-\(\frac{1}{2005}\)) : 3 = \(\frac{668}{2005}\)
1x4+2x5+...+277x280
= ( 1 + 1+ ... + 277) x ( 4 + 5 + .. + 280)
= 38503 x 39334 = 1514477002
mình mình k lại nha!
a)
\(P=\left(x^{14}-9x^{13}\right)-\left(x^{13}-9x^{12}\right)+\left(x^{12}-9x^{11}\right)-...+\left(x^2-9x\right)-\left(x-9\right)+1\)
\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+x^{11}\left(x-9\right)+...+x\left(x-9\right)-\left(x-9\right)+1\)
\(P\left(9\right)=1\)
b)
\(Q=\left(x^{15}-7x^{14}\right)-\left(x^{14}-7x^{13}\right)+\left(x^{13}-7x^{12}\right)-...-\left(x^2-7x\right)+\left(x-7\right)+2\)
\(=x^{14}\left(x-7\right)-x^{13}\left(x-7\right)+x^{12}\left(x-7\right)-...-x\left(x-7\right)+\left(x-7\right)+2\)
\(Q\left(7\right)=2\)
\(-\dfrac{1}{1\times4}-\dfrac{1}{4\times7}-\dfrac{1}{7\times10}-\dfrac{1}{10\times13}\)
\(=-\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+\dfrac{1}{10\times13}\right)\)
\(=-\dfrac{1}{3}\cdot\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}\right)\)
\(=-\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)\)
\(=-\dfrac{1}{3}\cdot\left(1-\dfrac{1}{13}\right)\)
\(=\dfrac{-1}{3}\cdot\dfrac{12}{13}\)
\(=\dfrac{-4}{13}\)
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