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a: =11/7(-3/7+4/11-4/7+7/11)=0
b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)
Lời giải:
a)
\(=\left(\frac{-3}{7}+\frac{4}{11}+\frac{-4}{7}+\frac{7}{11}\right):\frac{7}{11}=\left(\frac{-3-4}{7}+\frac{4+7}{11}\right):\frac{7}{11}=(-1+1):\frac{7}{11}=0\)
b)
Đặt biểu thức là $A$
\(-2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}-\frac{2}{97.99}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{97-95}{95.97}-\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}-\frac{2}{97.99}\)
\(=1-\frac{1}{97}-\frac{2}{97.99}=\frac{96.99-2}{97.99}\)
\(\Rightarrow A=\frac{1-48.99}{97.99}\)
Mình sửa lại chút.
\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(=\dfrac{1}{99.97}-\left\{\dfrac{1}{97.95}+\dfrac{1}{95.93}\right\}-\left\{\dfrac{1}{5.3}+\dfrac{1}{3.1}\right\}\)
\(=\dfrac{1}{99.97}-\dfrac{1}{95}.\left\{\dfrac{1}{97}+\dfrac{1}{93}\right\}-\dfrac{1}{3}.\left\{\dfrac{1}{5}+\dfrac{1}{1}\right\}\)
\(=\dfrac{1}{99.97}-\dfrac{1}{95}.\dfrac{190}{97.93}-\dfrac{1}{3}.\dfrac{6}{5}\)
\(=\dfrac{1}{99.97}-\dfrac{2}{97.93}-\dfrac{6}{15}\)
\(=\dfrac{1}{97}.\left\{\dfrac{1}{99}-\dfrac{2}{93}\right\}-\dfrac{2}{5}\)
\(=\dfrac{-35}{297693}-\dfrac{2}{5}\)
\(=\dfrac{-175-595386}{1488465}\)
\(=\dfrac{-595561}{1488465}\)
\(T=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)
\(T=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
Đặt \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)=\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{48}{97}\)
Thay \(A\) vào \(T\) ta có:\(T=\dfrac{1}{99\cdot97}-\dfrac{48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)
Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)
Đặt \(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\)
\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2B=1-\dfrac{1}{97}\)
\(2B=\dfrac{96}{97}\)
\(B=\dfrac{96}{97}:2\)
\(B=\dfrac{48}{97}\)
\(\Rightarrow A=\dfrac{1}{99.97}-\dfrac{48}{97}\)
\(A=\dfrac{1}{99.97}-\dfrac{48.99}{97.99}\)
\(A=\dfrac{1-48.99}{99.97}\)
\(A=-\dfrac{4751}{9603}\)
Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=-\dfrac{4751}{9603}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
a,
Đặt A = \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\right)\)
\(\Rightarrow\)2A= \(2.\left(\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{95}+...+\dfrac{1}{3}-1\right)\)
2A= \(2.\left(\dfrac{1}{99}-1\right)\)
\(\Rightarrow\) A = \(\dfrac{1}{99}-1=\dfrac{-98}{99}\)
b, \(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
= \(\dfrac{3.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5.\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}{5.\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)
= \(\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{5}{5}=1\)
Chúc bn hc tốt <3