Tính: 1+2-3-4+5+6-7-8+9+...........+305+306-307-308+309+310
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bài này dễ nhưng mất tg lắm k mình trước mình làm sau cho
Ta phân tích 308 thành 308 số 1 rồi nhóm lại
B=(1+307/2) + (1+306/3) +...+( 1+2/307) + (1+1/308) + 1=309/2+ 309/3+...+309/309
= 309( 1/2+1/3+...+ 1/308+1/309)=309A
A/B=1/309
Ta có :
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)
\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)
\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)
\(\frac{A}{B}=\frac{1}{309}\)
A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{308}\)+ \(\frac{1}{309}\)
B = \(\frac{308}{1}\)+ \(\frac{307}{2}\)+ \(\frac{306}{3}\)+\(\frac{3}{306}\) + \(\frac{2}{307}\)+ \(\frac{1}{308}\)
=> B = \(\frac{309-1}{1}\)+ \(\frac{309-3}{3}\)+... + ( 309 ... )
=> B = 309 + 309 . ( \(\frac{1}{2}\) + \(\frac{1}{3}\)+... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\)- \(\frac{1}{1}\)+ \(\frac{2}{2}\)+ ... + \(\frac{308}{308}\)+ \(\frac{309}{309}\)
=> B = 309 . ( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\))
=> \(\frac{A}{B}\)= \(\frac{1}{309}\)
ta có:1+2-3-4+5+6-7-8+9+.....+305+306-307-308+309+310
=1+(2-3-4+5)+(6-7-8+9)+....+(306-307-308+309)+310
=1+0+0+....+0+310
=1+310
=311
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