Theo đầu bài ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow2\cdot\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2017}\)
\(\Rightarrow x+1=2017\)
\(\Rightarrow x=2016\)

\(\frac{2}{6}\)\(+\frac{2}{12}\)\(+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}\div2\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)

\(\frac{1}{x+1}=\frac{1}{2017}\)

\(=>x+1=2017\)

\(=>x=2016\)

Chúc bạn học tốt Vu_anh_tuan ! 

Ta có: (x+2015)^2016>=0(với mọi x)

|y-2017|>=0(với mọi y)

Do đó, (x+2015)^2016+|y-2017|>=0(với mọi x,y)

mà (x+2015)^2016+|y-2017|=0

nên (x+2015)^2016=0                 và |y-2017|=0

      x+2015=0                              y-2017=0

      x=0-2015                              y=0+2017

      x=-2015                               y=2017

Vậy x=-2015 và y=2017 thì x,y thỏa mãn đề

1) Áp dụng tích chất dãy tỉ số bằng nhau ta có:

\(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y-x+y}{2015-2017}=\frac{2y}{-2}\)

\(=-y\)

\(\Rightarrow xy=-2016y;x+y=-2015y;\)

\(x-y=-2017y\)

\(\Rightarrow-2016y-xy=0\)

\(\Rightarrow y\left(-2016-x\right)=0\)

\(\Rightarrow\orbr{\orbr{\begin{cases}y=0\\-2016-x=0\end{cases}\Rightarrow}}\orbr{\begin{cases}y=0\\x=-2016\end{cases}}\)

\(+) \)\(y=0\Rightarrow0+x=-2015.0=0\Rightarrow x=0\)

\(+) \)\(x=-2016\Rightarrow-2016-y=-2017y\Rightarrow-2016\)

Vậy +) x=y=0

       +) x=-2016;y=1

2) Có: \(\frac{2x+2}{3}=\frac{x+1}{1,5};\frac{4z+2}{5}=\frac{z+0,5}{1,25};\frac{3y-1}{4}=\frac{y-\frac{1}{3}}{\frac{4}{3}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x+1}{1,5}=\frac{y-\frac{1}{3}}{\frac{4}{3}}=\frac{z+0,5}{1,25}=\frac{x+y+z+\left(1-\frac{1}{3}+0,5\right)}{1,5+\frac{4}{3}+1,25}=\frac{7+\frac{7}{6}}{\frac{49}{12}}=2\)

Suy ra: \(x+1=2.1,5=3\Rightarrow x=2\)

             \(y-\frac{1}{3}=2.\frac{4}{3}=\frac{8}{3}\Rightarrow y=3\)

            \(z+0,5=2.1,25=2,5\Rightarrow z=2\)

Vậy x=2;y=3;z=2.

Câu 1 :

Áp dụng t/c dãy TSBN ta có : \(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y+x-y}{2015+2017}=\frac{x}{2016}\)

\(\Rightarrow\frac{xy}{2016}=\frac{x}{2016}\)=> xy=x => xy-x=0 => x(y-1)=0 => x=0 hoặc y=1

+) Nếu x=0 => \(\frac{0+y}{2015}=\frac{0.y}{2016}\Rightarrow\frac{y}{2015}=0\Rightarrow y=0\)

+) Nếu y=1 => \(\frac{x+1}{2015}=\frac{x.1}{2016}\)=> 2016(x+1)=2015x => 2016x+2016 = 2015x => x=-2016

             Vậy ...

Câu 2 :

Áp dụng t/c dãy TSBN ta có : \(\frac{2x+2}{3}=\frac{3y-1}{4}=\frac{4z+2}{5}=\frac{6.\left(2x+2\right)+4.\left(3y-1\right)+3.\left(4z+2\right)}{3.6+4.4+5.3}\)

                                             \(=\frac{12\left(x+y+z\right)+14}{49}=\frac{12.7+14}{49}=2\)

Từ  \(\frac{2x+2}{3}=2\Rightarrow2x+2\Rightarrow6\Rightarrow2x=4\Rightarrow x=2\)

Tương tự tìm đc y=3 và z=2

            Vậy ...

=>(x-1)/2015 - 1 + (x-2(/2014 -1 = (x-3)/2013 -1 + (x-4)/2012 -1

=>(x-2016)*(1/2015+1/2014-1/2013-1/2012)=0

=>x=2016

Trừ 1 ở mỗi p/s,ta có:

\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)=\left(\frac{x-3}{2013}-1\right)+\left(\frac{x-4}{2012}-1\right)\)

\(\Leftrightarrow\left(\frac{x-2016}{2015}\right)+\left(\frac{x-2016}{2014}\right)=\left(\frac{x-2016}{2013}\right)+\left(\frac{x-2016}{2012}\right)\)

\(\Leftrightarrow\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

Vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)

=>x-2016=0

=>x=2016

Vậy..................

mình hỏi bài này :tìm số tự nhiên x biết :(x-2)^2014=(x-2)^2016

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

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