Giải pt:
\(\frac{x-1}{2015}+\frac{x-2}{2014}+...+\frac{x-2014}{2}+x=4030\)
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\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(1-\frac{2016}{1}\right)+\left(1-\frac{2017}{2}\right)+...+\left(1-\frac{4030}{2015}\right)\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
\(\Rightarrow x=2015\)
Không hiểu thì hỏi mình nhé! Thiên dâng bữa nay chăm chỉ đột xuất ta???
\(PT\Leftrightarrow\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=2017\)
Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành:
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)
\(\Rightarrow x=2018;y=2019;z=2020\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)
\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)
\(x=2018,y=2019,z=2020\)
Các bạn giúp mình giải bài này với
Giải PT sau:
\(\frac{2-x}{2013}-1=\frac{1-x}{2014}-\frac{x}{2015}\)
(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
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
\(\frac{x-1}{2015}+\frac{x-2}{2014}+...+\frac{x-2014}{2}+x=4030\)
\(\Leftrightarrow\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+...+\left(\frac{x-2014}{2}-1\right)+x-2016=0\)
\(\Leftrightarrow\frac{x-2016}{2015}+\frac{x-2016}{2014}+...+\frac{x-2016}{2}+x-2016=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}+1\right)=0\)
\(\Leftrightarrow x=2016\)
a Alibaba
đến bước này a phải nói
1/2015+ 1/2014 +.....+1/2+1 khác 0
=> x-2016=0=>x=2016 chứ
hơi lm tắt anh oi có ng` sẽ ko hiểu =))