Cho A = 1/2 + 1/3 + 1/4 + 1/5 + ............ + 1/308 + 1/309
B = 308/1 + 307/2 + 306/3 + ................... + 3/306 + 2/207 + 1/308
Tính A/B
Cho A= 1/2+1/3+1/4+1/5+...+1/308+1/309
B=308/1+307/2+306/3+305/4+...+3/306+2/307+1/308
Tính A/B
Cho:
A=1/2+1/3+1/4+1/5+...+1/308+1/309
B=308/1+307/2+306/3+...+3/306+2/307+1/308
Tính A/B ? ( Có cách làm )
cho A=1/2+1/3+1/4+...+1/308+1/309
B=308/1+307/2+306/3+...+3/306+2/307+1/308
tim A/B
A=1/2+1/3+1/4+...+1/308+1/309
B=308+307/2+306/3+...+3/306+2/307+1/308
Tính A/B.
bài này dễ nhưng mất tg lắm k mình trước mình làm sau cho
A=1/2+1/3+1/4+...+1/308+1/309
B=308/1+307/2+306/3+...+3/306+2/307+1/308
Tính A/B.
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
cho A =1/2 +1/3 +1/4 +.....+1/308+1/39
B = 308/1 +307/2 +.....+3/306 +2/307+1/308.Tính a+b
Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)
B=\(\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\). Tính \(\frac{A}{B}\)
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 . ( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\))
=> \(\frac{A}{B}\)= \(\frac{1}{309}\)
Lâu rồi bạn còn cần lời giải ko mình giải cho
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{308}+\frac{1}{309}\)
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+....+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
Tính\(\frac{\:A}{B}\)
Ta có :
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)
\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)
\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)
\(\frac{A}{B}=\frac{1}{309}\)
1/ Cho A=1/2+1/3+1/4+1/5+....+1/308+1/309; B=308/1+307/2+306/3+........+3/306+2/307+1/308. Tinh A/B
2/ Có hay không một số nguyên tố mà khi chia cho 12 dư 9 .Giải thích vì sao
3/ CMR: Trong 3 số nguyên tố lớn hơn 3,luôn tồn tại 2 số nguyên tố mà tổng hoặc hiệu của chúng chia hết cho 12
1, B = 308/1 + 307/2 + 306/3 + ... + 3/306 + 2/307 + 1/308
= ( 307/2 + 1 ) + ( 306/3 + 1 ) + ... + ( 3/306 + 1 ) + ( 2/307 + 1 ) + ( 1/308 + 1 ) + 1
= 309/2 + 309/3 + ... + 309/306 + 309/307 + 309/308 + 1
= 309 . ( 1/2 + 1/3 + ... + 1/306 + 1/307 + 1/308 + 1/309 )
= 309 . A
=> A/B = 1/309