tim x biet (x-1)^2017=4*(x-1)^2015
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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.
=>(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..................
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
(x - 1)2017 = 4(x - 1)2015
=> (x - 1)2017 - 4(x - 1)2015 = 0
=> (x - 1)2015.(x - 1)2 - 4(x - 1)2015 = 0
=> (x - 1)2015.[(x - 1)2 - 4] = 0
=> (x - 1)2015 = 0 hoặc (x - 1)2 - 4 = 0
=> x - 1 = 0 hoặc (x - 1)2 = 4
=> x = 1 hoặc x - 1 = 2 hoặc x - 1 = -2
=> x = 1 hoặc x = 3 hoặc x = -1
Vậy x = {-1;1;3}
(x-1)^2017=4*(x-1)^2015
(x-1)^2017/(x-1)^2015=4
(x-1)^2=4
(x-1)=2 hoặc -2
giải tiếp....................