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a: 17/200>17/314
b: 11/54=22/108<22/37
c: 141/893=3/19
159/901=3/17
mà 3/19<3/17
nên 141/893<159/901
tính chất trên gọi là tính chất bắc cầu, ta so sánh hai phân số với một số (phân số) thứ 3.
a) \(\dfrac{-1}{-4}\)=\(\dfrac{1}{4}>0\)
\(\dfrac{3}{-4}< 0\)
\(\Rightarrow\dfrac{1}{4}>\dfrac{3}{-4}hay\dfrac{-1}{-4}>\dfrac{3}{-4}\)
b) Ta có:
\(\dfrac{15}{17}=1-\dfrac{2}{17}\\ \)
\(\dfrac{25}{27}=1-\dfrac{2}{27}\\ \\ \)
Mà \(\dfrac{2}{17}>\dfrac{2}{27}\left(17< 27\right)\)
\(\Rightarrow1-\dfrac{2}{17}< 1-\dfrac{2}{27}\)hay \(\dfrac{15}{17}< \dfrac{25}{27}\)
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
Bài này có rất nhiều cách lm nhé!
Ta có : A = \(\dfrac{17^{18}+1}{17^{19}+1}\) => 17A = \(\dfrac{17^{19}+17}{17^{19}+1}\) = \(1+\dfrac{16}{17^{19}+1}\)
B = \(\dfrac{17^{17}+1}{17^{18}+1}\) => 17B = \(\dfrac{17^{18}+17}{17^{18}+1}\) = \(1+\dfrac{16}{17^{18}+1}\)
Vì \(\dfrac{16}{17^{19}+1}\) < \(\dfrac{16}{17^{18}+1}\) ( vì 1719 +1 > 1716+1 )
=> \(1+\dfrac{16}{17^{19}+1}\) < \(1+\dfrac{16}{17^{18}+1}\)
=> 17A < 17B
=> A < B ( vì 17 > 0)
Ta có :
\(A=\dfrac{17^{18}+1}{17^{19}+1}\)
17A= \(17\times\dfrac{17^{18}+1}{17^{19}+1}\)
\(17A=\dfrac{17^{19}+17}{17^{19}+1}\)
\(17A=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}\)
\(17A=\dfrac{17^{19}+1}{17^{19}+1}+\dfrac{16}{17^{19}+1}\)
\(17A=1+\dfrac{16}{17^{19}+1}\)
Lại có :
\(B=\dfrac{17^{17}+1}{17^{18}+1}\)
\(17B=17\times\dfrac{17^{17}+1}{17^{18}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}\)
\(17B=\dfrac{\left(17^{18}+1\right)+16}{17^{18}+1}\)
\(17B=\dfrac{17^{18}+1}{17^{18}+1}+\dfrac{16}{17^{18}+1}\)
\(17B=1+\dfrac{16}{17^{18}+1}\)
Mà : \(\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)
\(\Rightarrow1+\dfrac{16}{17^{19}+1}< 1+\dfrac{16}{17^{18}+1}\)
⇒ A < B
Vậy A < B
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5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)
\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)
\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)
\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)
Bài 1:
a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)
\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)
c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)
\(\dfrac{-7}{-17}=\dfrac{7}{17}\)
Vì 7>6 nên \(\dfrac{-7}{-17}>\dfrac{6}{17}\)
\(\dfrac{-7}{-17}=\dfrac{7}{17}\)
\(\dfrac{6}{17}\) giữ nguyên
Vì \(7>6\)
\(\Rightarrow\dfrac{-7}{-17}>\dfrac{6}{11}\)