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a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}..1\frac{1}{99}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{100}{99}=\frac{2.2.3.3.4.4...10.10}{1.3.2.4.3.5...9.11}=\frac{\left(2.3.4...10\right)\left(2.3.4...10\right)}{\left(1.2.3...9\right)\left(3.4.5...11\right)}\)
\(\frac{10.2}{1.11}=\frac{20}{11}\)
b) \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}\)
\(=\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(1.2.3.4.5\right).\left(3.4.5.6.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{1.7}{6.2}=\frac{7}{12}\)
c) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)+\left(-\frac{98}{97}+\frac{1}{97}\right)=1-1=0\)
d) \(3\frac{1}{11}.\frac{27}{36}.1\frac{6}{7}.2\frac{4}{9}=\frac{34}{11}.\frac{3}{4}.\frac{13}{7}.\frac{22}{9}=\frac{34.3.13.22}{11.4.7.9}=\frac{34.13}{11.2.7.3}=\frac{442}{462}=\frac{221}{231}\)
a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
a) x + 1/5 + x + 1/7 = x + 1/9
<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9
<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9
<=> 12/35x + 12/35 = 1/9x + 1/9
<=> 12/35x + 12/35 - 1/9x = 1/9
<=> 73/315x + 12/35 = 1/9
<=> 73/315x = 1/9 - 12/35
<=> 73/315x = -73/315
<=> x = 73/315 : -73/315 = -1
=> x = -1
b) làm tương tự
\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)
\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)
\(\Rightarrow x=1\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)
\(B=\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1\)
\(B=\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}\)
\(B=100\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)\)
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 :
\(B=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)
\(B=1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)\)
\(B=\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)
\(B=100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}\right)}=100\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{99}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{98}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Ta có:\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
\(B=\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}\)
\(B=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(B=99+\frac{98}{2}+...+\frac{1}{99}\)
\(B=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
(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) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)
\(=\frac{98}{98}-\frac{97}{97}=1-1=0\)