\(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
Đặt x2 - 2x + 2 = t, ta có:
\(\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}\)
<=> \(\frac{\left(t+1\right)\left(t+2\right)}{t\left(t+1\right)\left(t+2\right)}+\frac{2t\left(t+2\right)}{t\left(t+1\right)\left(t+2\right)}=\frac{6t\left(t+1\right)}{t\left(t+1\right)\left(t+2\right)}\)
=> \(\left(t+1\right)\left(t+2\right)+2t\left(t+2\right)=6t\left(t+1\right)\)
<=> t2 + 2t + t + 2 + 2t2 + 4t = 6t2 + 6t
<=> 3t2 + 7t - 6t2 - 6t + 2 = 0
<=> - 3t2 + 3t - 2t + 2 = 0
<=> 3t(1- t) + 2(1 - t) = 0
<=> (1 - t)(3t + 2) = 0
<=> 1 - t = 0 hoặc 3t + 2 = 0
+) 1 - t = 0
<=> 1 - x2 + 2x - 2 = 0
<=> - x2 + 2x - 1 = 0
<=> - (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
+) 3t + 2 = 0
<=> 3(x2 - 2x + 2) + 2 = 0
<=> 3x2 - 6x + 6 + 2 = 0
<=> 3x2 - 6x + 3 + 5 = 0
<=> 3(x2 - 2x + 1) + 5 = 0
<=> 3(x - 1)2 + 5 = 0
mà \(3\left(x-1\right)^2+5\ge5>0\)
=> Phương trình vô nghiệm
Vậy tập nghiệm của phương trình là S = {1}
Có : \(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Leftrightarrow\) \(\left(\frac{x+2011}{2013}+1\right)\)+ \(\left(\frac{x+2012}{2012}+1\right)\) - \(\left(\frac{x+2010}{2014}+1\right)\) - \(\left(\frac{x+2013}{2011}+1\right)=0\)
\(\Leftrightarrow\) \(\frac{x+2025}{2013}+\frac{x+2025}{2012}-\frac{x+2025}{2014}-\frac{x+2025}{2011}\)
\(\Leftrightarrow\) \(\left(x+2025\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2015}\right)\)=0
\(\Leftrightarrow\)\(x+2025=0\) ( vì \(\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)\ne0\))
\(\Rightarrow\)\(x=-2025\)
Vậy \(x=-2025\)
ây da nhầm rồi để mk làm lại cho kết quả là -4025 mới phải
Sửa lại nè !
Có : \(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\)\(\frac{x+2010}{2014}+\frac{2013}{2011}\)
\(\Leftrightarrow\) \(\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)\)- \(\left(\frac{x+2010}{2014}+1\right)-\left(\frac{x+2013}{2011}+1\right)=0\)
\(\Leftrightarrow\) \(\frac{x+4024}{2013}+\frac{x+4024}{2012}-\) \(\frac{x+4024}{2014}-\frac{x+4024}{2011}\)=0
\(\Leftrightarrow\)\(\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)\)=0
\(\Leftrightarrow\) \(x+4024=0\) ( vì \(\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)\ne0\)
\(\Leftrightarrow\) \(x=-4024\)
Vậy \(x=-4024\)
Đặt x^2-2x+3=t (*)
\(\frac{1}{t-1}+\frac{2}{t}-\frac{6}{t+1}=0\Leftrightarrow\frac{t^2+t+2t^2-2-6t^2+6t}{t\left(t^2-1\right)}=\frac{-3t^2+7t-2}{t\left(t^2-1\right)}=\frac{-\left(t-2\right)\left(3t-1\right)}{t\left(t^2-1\right)}=0\Rightarrow\left[\begin{matrix}t=2\\t=\frac{1}{3}\end{matrix}\right.\)
với t=2 => x=1
với t=1/3 (*) vô nghiệm
