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\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)
\(B=\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1\)
\(B=\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}\)
\(B=100\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)\)
Ta có: \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}=\frac{1}{100}\)
Vậy...
P/s: Hoq chắc
#)Giải :
\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)
\(B=1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)\)
\(B=\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)
\(B=100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}=100\)
\(B=\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}\)
\(B=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(B=99+\frac{98}{2}+...+\frac{1}{99}\)
\(B=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
(số hạng 99 chia thảnh 99 số 1 cộng vào từng phân số còn dư 1 số 1 để ngoài)
\(B=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)
\(B=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{B}{A}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}\)
\(\Rightarrow\frac{B}{A}=100\)
b/a = 100. Nếu k đúng cho mình, Mình sẽ trình bày cách làm cho bạn.
a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.........1\frac{1}{99}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}......\frac{100}{99}\)
\(=\frac{\left(2.2\right).\left(3.3\right).\left(4.4\right).\left(5.5\right)....\left(10.10\right)}{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right).....\left(9.11\right)}\)
\(=\frac{\left(2.3.4.5...10\right).\left(23.4.5....10\right)}{\left(1.2.3.4...9\right).\left(3.4.5.6....11\right)}=\frac{10}{1}.\frac{2}{11}=\frac{20}{11}\)
b) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)
\(=\frac{98}{98}-\frac{97}{97}=1-1=0\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{99}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{98}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Mình biết trả lời câu này nhưng mình đang rất bận ! Có gì vào nhắn tin cho mình tối mình giải cho !
\(B=\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):...:\left(\frac{1}{98}-1\right):\left(\frac{1}{99}-1\right):\left(\frac{1}{100}-1\right)\)
\(B=\frac{1}{2}:\frac{2}{3}:\frac{3}{4}:....:\frac{97}{98}:\frac{98}{99}:\frac{99}{100}\)
\(B=\frac{1}{2}\times\frac{3}{2}\times\frac{4}{3}\times...\times\frac{98}{97}\times\frac{99}{98}\times\frac{100}{99}\)
\(B=\frac{1\times3\times4\times...\times98\times99\times100}{2\times2\times3\times...\times97\times98\times99}\)
\(B=\frac{1\times100}{2\times2}\)
\(B=\frac{100}{4}\)
\(B=25\)
HOK TOT
Đặt mẫu là A :
=> A = \(\frac{1}{99}\)+ \(\frac{2}{98}\)+ \(\frac{3}{97}\)+...+ \(\frac{98}{2}\)+ \(\frac{99}{1}\)
= ( \(\frac{1}{99}\)+ 1 ) + ( \(\frac{2}{98}\)+ 1 ) + ( \(\frac{3}{97}\)+ 1 ) +...+ ( \(\frac{98}{2}\)+ 1 ) + ( \(\frac{99}{1}\)+1 ) - 99 ( Vì ta đã cộng với 99 số 1 rồi nên phải trừ 99 )
= \(\frac{100}{99}\)+ \(\frac{100}{98}\)+ \(\frac{100}{97}\)+...+ \(\frac{100}{2}\)+ \(\frac{100}{1}\)- 99
= \(\frac{100}{99}\)+ \(\frac{100}{98}\)+ \(\frac{100}{97}\)+...+ \(\frac{100}{2}\)+ ( \(\frac{100}{1}\)- 99 )
= \(\frac{100}{99}\)+ \(\frac{100}{98}\)+ \(\frac{100}{97}\)+...+ \(\frac{100}{2}\)+ \(\frac{100}{100}\)
= 100.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{100}\))
=> Tử trên mẫu là :
\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{100}\)
_______________________________________
100.(\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{100}\))
=\(\frac{1}{100}\)
1) \(\left(x-1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\right)=0\)
mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
2) \(\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-5}{95}-1=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{95}=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
x - 100 = 1
x = 101
1x100/99x99/98x98/97x.....x3/2x2
1x100
100
\(1:\frac{99}{100}:\frac{98}{99}:\frac{97}{98}:.........:\frac{2}{3}:\frac{1}{2}\)
\(=1.\frac{100}{99}.\frac{99}{98}.\frac{98}{97}......\frac{3}{2}.\frac{2}{1}\)
\(=\frac{1.100.99.98....3.2}{99.98.97......2.1}\)
\(=100\)