1) tinh R2
a) \(\dfrac{2R_2}{\left(\dfrac{3R_2}{3+R_2}\right)}\)
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a/ \(\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.36\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+2+3+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
\(=0\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{3}{4}\right):\left(4+\dfrac{3}{4}-3-\dfrac{1}{2}\right)\)
\(=\dfrac{4}{3}:\left(1+\dfrac{1}{4}\right)=\dfrac{4}{3}:\dfrac{5}{4}=\dfrac{16}{15}\)
a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)
b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)
c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)
=-520/7
Giải:
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)
\(=\dfrac{1.2.3...99}{2.3.4...100}\)
\(=\dfrac{1}{100}\)
Vậy ...
e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)
=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)
=\(-\dfrac{516}{7}\)
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)
=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)
=\(\dfrac{119}{240}\)
a: \(=\dfrac{7+3}{6}\cdot\dfrac{1}{2}-2:\dfrac{7+3}{6}\)
\(=\dfrac{10}{12}-2\cdot\dfrac{6}{10}\)
\(=\dfrac{5}{6}-\dfrac{6}{5}=\dfrac{25-36}{30}=-\dfrac{11}{30}\)
b: \(=\left|\dfrac{10}{5}-\dfrac{2}{5}\right|\cdot\dfrac{1}{27}-\dfrac{3}{5}+1\)
\(=\dfrac{8}{5}\cdot\dfrac{1}{27}-\dfrac{3}{5}+1\)
\(=\dfrac{8}{135}-\dfrac{81}{135}+\dfrac{135}{135}=\dfrac{62}{135}\)
đề hồi nãy thì dấu chia còn đề này thì nhân
rốt cuộc là nhân hay chia
\(\dfrac{2R_2}{\left(\dfrac{3R_2}{3+R_2}\right)}=\dfrac{2R_2\left(3+R_2\right)}{3R_2}=\dfrac{6}{3}+\dfrac{2R_2}{3}=2+\dfrac{2}{3}R_2\)
Vì đề bài không cho biểu thức trên có giá trị bằng bao nhiêu nên mình chỉ làm như vậy