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\(\text{K - 2016 = }\frac{\text{1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ... + ( 1 + 2 + 3 + ... + 2017 )}}{\text{2017 x 1 + 2016 x 2 + 2015 x 3 + ... + 2 x 2016 + 1 x 2017}}\)
\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)
\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)
mấy câu kia áp dụng là dc!
chi tinh ra thi la 25/6 nha em. em kiem tra ket qua nhe!!!!!!! ^_^
Bạn có thể cho mình biết cách giải được không vậy bạn.
a ) 23/33 > 22/33 = 2/3 = 6/9 > 6/71
Vậy ta có 6/71 ; 2/3 ; 23/33
b ) 1/2 = 4/8 ; 3/4 = 6/8
Vậy ta có 1/2 ; 5/8 ; 3/4
c ) 8/9 > 8 / 11 ; 8/9 > 9/11 ; 9/11 > 8/11
Vậy ta có 8/11 ; 9/11 ; 8/9
\(5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}\times4\dfrac{1}{2}-2\times2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}\times\dfrac{9}{2}-2\times\dfrac{7}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left[\dfrac{7}{3}\times\left(\dfrac{9}{2}-2\right)\right]:\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\left(\dfrac{7}{3}\times\dfrac{5}{2}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}\cdot\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{59}{15}-\dfrac{50}{15}\)
\(=\dfrac{9}{15}\)
\(=\dfrac{3}{5}\)
\(Toru\)
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\cdot\left(2x+3\right)}\)
\(=\frac{1}{3}\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{\left(2x+1\right)\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x+3}{3\left(2x+3\right)}-\frac{3}{3\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x+3-3}{3\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x}{6x+9}\right)\)
\(=\frac{2x}{3\left(6x+9\right)}=\frac{2x}{18x+27}\)
\(\frac{3}{5\times7}+\frac{3}{7\times9}+...+\frac{3}{117\times119}\)
\(=\frac{3}{2}\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{117\times119}\right)\)
\(=\frac{3}{2}\times\left(\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+...+\frac{119-117}{117\times119}\right)\)
\(=\frac{3}{2}\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{117}-\frac{1}{119}\right)\)
\(=\frac{3}{2}\times\left(\frac{1}{5}-\frac{1}{119}\right)=\frac{171}{595}\)