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\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
\(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
\(=\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)
\(=\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{7}-\dfrac{1}{11}\)
\(=\dfrac{11}{77}-\dfrac{7}{77}\)
\(=\dfrac{4}{77}\)
Giải thích các bước giải:
(7^108 x 50 -7^108) : 7^110
= [ 7^108 x (50-1) ] : 7^110
= [ 7^108 x 49] : 7^110
= ( 7^108 x 7^2) : 7^110
= 7^110 - 7^110
=1
=\(\frac{7^{108}\cdot50-7^{108}}{7^{110}}\)
=\(\frac{7^{108}\cdot50-7^{108}\cdot1}{7^{110}}\)
=\(\frac{7^{108}\cdot\left(50-1\right)}{7^{110}}\)
=\(\frac{7^{108}\cdot49}{7^{110}}\)
=\(\frac{7^{108}\cdot7^2}{7^{110}}\)
=\(\frac{7^{108+2}}{7^{110}}\)
=\(\frac{7^{110}}{7^{110}}\)=1