\(\dfrac{73}{100}\) - \(\dfrac{47}{30}\) : \(\dfrac{34}{78}\) =?
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\(\dfrac{5}{17}\times\dfrac{34}{25}=\dfrac{5\times17\times2}{17\times5\times5}=\dfrac{2}{5}\)
\(\dfrac{5}{17}\times\dfrac{34}{25}=\dfrac{5\times34}{17\times25}=\dfrac{170}{425}=\dfrac{2}{5}\)
\(=\dfrac{-8}{27}\cdot81+\dfrac{9}{16}\cdot32\)
=-24+18
=-6
1,
Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)
\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)
2,
Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)
3,
Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)
\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)
4,
\(\dfrac{48}{47}>1>\dfrac{84}{85}\)
Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)
Ta đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(a=bk\)
\(c=dk\)
Ta có:
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2\times k^2+b^2}{d^2\times k^2+d^2}=\dfrac{b^2\times\left(k^2+1\right)}{d^2\times\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)
=> \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
=> đpcm
Áp dụng công thức tỉ lệ phân số ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{ac}{bd}\)
\(\dfrac{6}{7}+\dfrac{5}{8}=\dfrac{48}{56}+\dfrac{35}{56}=\dfrac{83}{56}\)
\(\dfrac{36}{12}-\dfrac{5}{3}=\dfrac{9}{3}-\dfrac{5}{3}=\dfrac{4}{3}\)
\(\dfrac{6}{7}+\dfrac{5}{8}=\dfrac{48}{56}+\dfrac{35}{56}=\dfrac{83}{56}\\\dfrac{36}{12}+\dfrac{5}{3}=\dfrac{36}{12}+\dfrac{20}{12} =\dfrac{56}{12}=\dfrac{14}{3}\)
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\(\dfrac{\dfrac{5}{7}+\dfrac{5}{9}+\dfrac{5}{11}}{\dfrac{25}{7}+\dfrac{25}{9}+\dfrac{25}{11}}\cdot\dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}\)
=\(\dfrac{5\cdot\left(\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}\right)}{25\left(\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}\right)}\cdot\dfrac{4\left(1+\dfrac{1}{73}+\dfrac{1}{115}\right)}{5+\dfrac{5}{73}-\dfrac{5}{115}}\)
=\(\dfrac{5}{25}\cdot\dfrac{4\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}{5\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}\)
=\(\dfrac{1}{5}\cdot\dfrac{4}{5}\)=\(\dfrac{4}{25}\)
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