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\(\frac{2016+x}{x}+\frac{2520+x}{x}+\frac{3024+x}{x}=2523\)
\(\Leftrightarrow\frac{2016+x+2520+x+3024+x}{x}=2523\)
\(\Leftrightarrow\frac{7560+3x}{x}=\frac{2523x}{x}\)Khử mẫu : \(7560+3x=2523x\)
\(\Leftrightarrow7560=2520x\Leftrightarrow x=3\)
\(\Rightarrow\frac{2016}{x}+\frac{x}{x}+\frac{2520}{x}+\frac{x}{x}+\frac{3024}{x}+\frac{x}{x}=2523\)
\(\Rightarrow\frac{2016}{x}+\frac{2520}{x}+\frac{3024}{x}=2520\)
\(\Rightarrow\frac{1}{x}\left(2016+2520+3024\right)=2520\)
\(\Rightarrow x=3\)
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
Ta phân tích :13=2+3+4
=> 1/2+1/3+1/4=13/12.
=> Ta có : 1/2016-x=1/2
=> Ta có 2016-x=2 =>x=2014
Hãy tích cho tui đi
vì câu này dễ mặc dù tui ko biết làm
Yên tâm khi bạn tích cho tui
Tui sẽ ko tích lại bạn đâu
THANKS
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}...\frac{2018}{2016}.\frac{2019}{2017}\\ =\frac{3.4.5...2018.2019}{1.2.3...2016.2017}\\ =\frac{2018.2019}{2}=1009.2019\)
Ta có: \(1\times\frac{1}{15}\times1\frac{1}{16}\times1\frac{1}{17}\times......\times1\frac{1}{2016}\times1\frac{1}{2017}\)
\(=\frac{1}{15}\times\frac{17}{16}\times\frac{18}{17}\times......\times\frac{2017}{2016}\times\frac{2018}{2017}\)
\(=\frac{1}{15}\times\frac{2018}{16}\)
\(=\frac{1009}{8\times15}\)
\(=\frac{1009}{120}\)
Mình nghĩ đề bài của bạn bị nhầm ở chỗ \(1\frac{1}{15}\)thành \(1\times\frac{1}{15}\)
Nhưng không sao bạn ạ
Vẫn giải được
Sửa đề \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\Leftrightarrow\frac{1}{x+1}=\frac{1}{4032}\)
\(\Leftrightarrow x+1=4032\Rightarrow x=4031\)
\(\frac{2016+x}{x}+\frac{2520+x}{x}+\frac{3024+x}{x}=\frac{2016}{x}+\frac{x}{x}+\frac{2520}{x}+\frac{x}{x}+\frac{3024}{x}+\frac{x}{x}\)
\(=\frac{2016+2520+3024}{x}+3\)
\(=\frac{7560}{x}+3\)
\(\frac{7560}{x}+3=2523\)
\(\frac{7560}{x}=2520\)
\(x=\frac{7560}{2520}\)
\(x=3\)