Limx->1 (\(\frac{2017}{1-x^{2017}}-\frac{2018}{1-x^{2018}}\) )
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26 tháng 7 2019
\(\frac{x-2017}{2018}-\frac{x-2018}{2017}=\frac{2017}{x-2018}-\frac{2018}{x-2017}\)
\(\Leftrightarrow\)\(\frac{2017\left(x-2017\right)-2018\left(x-2018\right)}{2017.2018}=\frac{2017\left(x-2017\right)-2018\left(x-2018\right)}{\left(x-2017\right)\left(x-2018\right)}\)
Do \(2017\left(x-2017\right)-2018\left(x-2018\right)\ne0\) nên \(\left(x-2017\right)\left(x-2018\right)=2017.2018\)
\(\Leftrightarrow\)\(x^2-4035x+2017.2018=2017.2018\)
\(\Leftrightarrow\)\(x\left(x-4035\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(l\right)\\x=4035\left(n\right)\end{cases}}\)
Vậy x = 4035
12 tháng 5 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=41-2\)
\(\Leftrightarrow x=39\)
7 tháng 8 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
CV
19 tháng 4 2019
đặt x-2016=a
y-2017=b
z-2018=c
ta có\(\frac{1}{\sqrt{a}}-\frac{1}{a}+\frac{1}{\sqrt{b}}-\frac{1}{b}+\frac{1}{\sqrt{c}}-\frac{1}{c}=\frac{3}{4}\)
=>\(\left(\frac{1}{\sqrt{a}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{2}\right)^2=0\)
=>\(a=b=c=4\)
còn lại tự lm nốt