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CN
0
20 tháng 4 2018
Bài 3 :
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-4}{2013}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1-2016}{2016}+\frac{x-2-2015}{2015}=\frac{x-3-2014}{2014}+\frac{x-4-2013}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
Nên \(x-2017=0\)
\(\Rightarrow\)\(x=2017\)
Vậy \(x=2017\)
Chúc bạn học tốt ~
20 tháng 4 2018
Bài 1 :
\(\left(8x-5\right)\left(x^2+2014\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{5}{8}\)
Chúc bạn học tốt ~
BH
21 tháng 2 2019
\(\frac{2}{\left(x+3\right)\left(x+1\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}-\frac{2}{x+3}+\frac{2}{x+3}-\frac{2}{x+5}+\frac{2}{x+5}-\frac{2}{x+7}=\frac{2}{9}\)
\(\frac{2}{x+1}-\frac{2}{x+7}=\frac{2}{9}\\ \Rightarrow\frac{2x+14-2x-2}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\\ \Rightarrow\frac{12}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}=\frac{12}{54}\)
\(\Rightarrow\left(x+1\right)\left(x+7\right)=54\\ \Rightarrow x^2+8x-54=0\Rightarrow x=-4\pm\sqrt{70}\)
D
2
27 tháng 2 2018
\(\frac{x-3}{2015}+\frac{x-2}{2016}=\frac{x-2016}{2}+\frac{x-2015}{3}\)
\(\Leftrightarrow\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-2}{2016}-1\right)=\left(\frac{x-2016}{2}-1\right)+\left(\frac{x-2015}{3}-1\right)\)
\(\frac{x-2018}{2015}+\frac{x-2018}{2016}-\frac{x-2018}{2}-\frac{x-2018}{3}=0\)
\(\Leftrightarrow\left(x-2018\right).\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}< 0\)
nên x - 2018 = 0
,<=> x = 2018
Vậy phương có 1 nghiệm là x = 2018
27 tháng 2 2018
pt <=> (x-3/2015 - 1) + (x-2/2016 - 1) = (x-2016/2 - 1) + (x-2015/3 - 1)
<=> x-2018/2015 + x-2018/2016 = x-2018/2 + x-2018/3
<=> x-2018/2 + x-2018/3 - x-2018/2015 - x-2018/2016 = 0
<=> (x-2018).(1/2+1/3-1/2015-1/2016) = 0
<=> x-2018 = 0 ( vì 1/2+1/3-1/2015-1/2016 > 0 )
<=> x=2018
Tk mk nha
23 tháng 3 2019
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)
\(\Leftrightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\)
Vậy x = 0
23 tháng 3 2019
\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)
\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow-6x+3=8\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy ...