tính: \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)
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
tính: \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)
ĐặtA= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)
\(\frac{1}{A}=1\div\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\right)\)
=> \(\frac{1}{A}=2+3+4+...+308+309\)
=>Ta có: Số các số hạng là:(309-2)/1+1=308(số hạng)
Tổng của\(\frac{1}{A}\)là:\(\frac{\left(309+2\right).308}{2}\)=47894
=> \(\frac{1}{A}=47894\)
=>\(A=\frac{1}{47894}\)
Chúc bạn học tốt!
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{308}+\frac{1}{309}\)
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+....+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
Tính\(\frac{\:A}{B}\)
Ta có :
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)
\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)
\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)
\(\frac{A}{B}=\frac{1}{309}\)
Cho: \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\)
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
Tính \(\frac{A}{B}\)
Cố lên nha các bạn!!! Mình sẽ tick cho bạn nào tính nhanh nhất
\(B=308/1+307/2+306/3+...+1/308 \)
\(B=308+307/2+306/3+...+1/308\) chia số 308 thành 308 số 1
B=307/2+1+306/3+1+...+1/308+1+1
B=309/2+309/3+309/4+...+309/308+309/309
B=309(1/2+1/3+1/4+...+1/309)=309A
Suy ra A/B=1/309
=(1/2+1/31/4...1/307/1/3081/309)/(309-1/1+309-2/2+...+309-307/307+309-308/308)
=(1/21/31/4...1/3071/3081/309)/(309/1-1+309/2-1+...+309/307-1+309/308-1)
=(........................................)/(309/309309/2309/3...309/307+309/308)
=(........................................)/[309x(1/309+1/308+...+1/41/31/2)]
Thấy tử và mẫu giống nhau thì ta rút:
=1/309
Tìm \(\frac{A}{B}\) biết A= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)
B= \(\frac{308}{1}+\frac{307}{2}+...+\frac{2}{307}+\frac{1}{308}\)
cho a) A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+....+\(\frac{1}{309}\), B = \(\frac{308}{1}\)+ \(\frac{307}{2}\)+....+ \(\frac{2}{307}\)+ \(\frac{1}{308}\). Tính \(\frac{A}{B}\)
b) \(\frac{7}{10.11}\)+ \(\frac{7}{11.12}\)+ \(\frac{7}{12.13}\)+....+ \(\frac{7}{69.70}\)
a. Ta có :
B = 308/1 + 307/2 +306/3+....+1/308
B = (1+1+....+1) + 307/2 + ....+ 1/308
B = (1 + 307/2) + (1+306/3) + ...+ (1+ 1/308) + 1
B = 309/2 + 309/3 + ....+ 309/308 + 309/309
B = 309.(1/2 + 1/3 + ....+1/309)
Vậy A/B: 1/2 + 1/3 + ... + 1/309 / 308/1 + 307/2 +....+ 2/307+1/308
A/B = 1/2 + 1/3 +... + 1/309 / 309.(1/2 + 1/3 + ....+1/309)
A/B = 1/309
b.7/10.11 + 7/11.12 + .... +7 /69.70
= 7. (1/10.11+1/11.12 + ...+ 1/69.70)
= 7.(1/10-1/11+1/11-1/12+....+1/69-1/70)
= 7.(1/10 - 1/70)
= 7. 3/35
= 3/5
Tính giá trị biểu thức:
A = \(2\frac{1}{309}\times\frac{1}{785}-\frac{1}{103}\times3\frac{784}{785}-\frac{4}{309\times785}+\frac{4}{103}\)
B = \(4\frac{1}{113}\times\frac{1}{371}-\frac{2}{113}\times5\frac{370}{371}-\frac{3}{113\times371}-\frac{4}{371}\)
C = \(x^3-31x^2-32x+7\) tại x = 32
D = \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)-x^2\)tại x = a + b + c
A = \(2\frac{1}{309}\times\frac{1}{785}-\frac{1}{103}\times3\frac{784}{785}-\frac{4}{309\times785}+\frac{4}{103}\)
B = \(4\frac{1}{113}\times\frac{1}{371}-\frac{2}{113}\times5\frac{370}{371}-\frac{3}{113\times371}-\frac{4}{371}\)
C = \(x^3-31x^2-32x+7\) tại \(x=32\)
D = \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)-x^2\)tại \(x=a+b+c\)
1/\(\left\{{}\begin{matrix}\frac{5}{x+3}-\frac{9}{y-2}=100\\\frac{3}{x+3}+\frac{7}{y-2}=308\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}\frac{3x}{x+1}-\frac{2y}{y+4}=4\\\frac{2x}{x+1}-\frac{5y}{y+4}=9\end{matrix}\right.\)
MỌI NGƯỜI GIÚP MÌNH VỚI !!!