Giải pt: \(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
\(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
Hãy giải pt này
\(PT\Leftrightarrow\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=2017\)
Giải phương trình \(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
\(\frac{x-3}{2014}\)-\(\frac{x-2}{2015}\)=\(\frac{x-1}{1008}\)+\(\frac{x}{2017}\)-1
gia sai đề rồi kía
tui làm ròi nên biết
\(\frac{x-3}{2014}-\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
=> \(\frac{x-3}{2014}-\frac{x-2}{2015}-2=\frac{x-1}{1008}+\frac{x}{2017}-1-2\)
=> \(\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
=> \(\frac{x-2017}{2014}-\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
=> \(\frac{x-2017}{2014}-\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}\)
=> \(\left(x-2017\right)\left(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
Vì \(\frac{1}{1008}>\frac{1}{2014}\) nên \(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}< 0\)
=> x-2017=0
=>x=2017
\(\frac{x}{2012}+\frac{x-1}{2013}+\frac{x-2}{2014}-\frac{x-3}{2015}=\frac{x-4}{1008}\)
\(\frac{x}{2012}+\frac{x-1}{2013}+\frac{x-2}{2014}-\frac{x-3}{2015}=\frac{x-4}{1008}\)
\(\Leftrightarrow\left(\frac{x}{2012}+1\right)+\left(\frac{x-1}{2013}+1\right)+\left(\frac{x-2}{2014}+1\right)-\left(\frac{x-3}{2015}+1\right)=\frac{x-4}{1008}+2\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x-4+1008.2}{1008}\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x+2012}{1008}\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}-\frac{x+2012}{1008}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\ne0\)
\(\Rightarrow x+2012=0\)\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\)
Chu Công Đứcbạn làm kết quả đúng nhưng trình bày sai rồi vế phải bạn cộng 2 nhưng vế trái bạn cộng 4???
Nhưng nếu vế phải +2 thì kết quả có thay đổi ko vậy ạ
Giải các pt sau:
a, (4x-1)(x+5)=(2x-3)^2
b, x(x+1)(x+2)(x+3)=24
c, x^2-2x+1=3x(x-1)
d,\(\frac{x+1}{2017}+\frac{x+2}{2015}=\frac{x+2014}{3}+\frac{x+2013}{4}\)
b) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x=t\) ta có:
\(t\left(t+2\right)-24=0\)
\(\Leftrightarrow\)\(t^2+2t-24=0\)
\(\Leftrightarrow\)\(\left(1-4\right)\left(1+6\right)=0\)
đến đây bn giải tiếp
Giải phương trình: \(\frac{x}{2017}+\frac{x+1}{2016}=\frac{x+2}{2015}+\frac{x+3}{2014}\)
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
Ta có: \(\frac{x}{2017}+\frac{x+1}{2016}=\frac{x+2}{2015}+\frac{x+3}{2014}\)
\(\Leftrightarrow\frac{x}{2017}+1+\frac{x+1}{2016}+1=\frac{x+2}{2015}+1+\frac{x+3}{2014}+1\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}-\frac{x+2017}{2015}-\frac{x+2017}{2014}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\)
nên x+2017=0
hay x=-2017
Vậy: S={-2017}
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x}{2016}+\frac{x-1}{2017}\)
Tìm x của pt bằng cach giải ko cần qui đồng.
Bạn nào trả lời đúng mình tick cho. Thanks nhiều! (^ - ^)
(x+2/2014)+1 + (x+1/2015)+1 = (x+2016)+1 + (x-1/2017)+1
(x+2016/2014) + (x+2016/2015) - (x+2016/2016) - (x-2016/2017)=0
=>(x+2016)(1/2014+1/2015-1/2016-1/2017)
vì 1/2014+1/2015-1/2016-1/2017 luôn khác 0 => x+2016=0
=> x=-2016
Giải pt:
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành:
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)
\(\Rightarrow x=2018;y=2019;z=2020\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)
\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)
\(x=2018,y=2019,z=2020\)
ĐK : \(\hept{\begin{cases}x>2014\\y>2015\\z>2016\end{cases}}\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2014}-1}{x-2014}+\frac{1}{4}-\frac{\sqrt{y-2015}-1}{y-2015}+\frac{1}{4}-\frac{\sqrt{z-2016}-1}{z-2016}=0\)
\(\Leftrightarrow\frac{x-2010-4\sqrt{x-2014}}{4\left(x-2014\right)}+\frac{y-2011-4\sqrt{y-2015}}{4\left(y-2015\right)}+\frac{z-2012-4\sqrt{z-2016}}{4\left(x-2014\right)}=0\)
\(\Leftrightarrow\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}+\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}+\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}=0\)( 1 )
Mà \(\hept{\begin{cases}\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}\ge0\forall x>2014\\\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}\ge0\forall y>2015\\\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}\ge0\forall z>2016\end{cases}}\)( 2 )
Từ ( 1 ) và ( 2 ) => \(\hept{\begin{cases}\left(2-\sqrt{x-2014}\right)^2=0\\\left(2-\sqrt{y-2015}\right)^2=0\\\left(2-\sqrt{z-2016}\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}\sqrt{x-2014}=2\\\sqrt{y-2015}=2\\\sqrt{z-2016}=2\end{cases}}\)<=>\(\hept{\begin{cases}x=2018\\y=2019\\z=2020\end{cases}}\)( tmđk )
Vậy ( x ; y ; z ) = ( 2018 ; 2019 ; 2020 )
Tìm x biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4029}{2014}+\frac{4030}{2015}\)