Câu hỏi của Nam Phạm An

Question

$\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}$

Y · Accepted Answer

+ Đặt $t=x^2-3x+3$ thì pt đã cho trở thành :

$\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}$

$\Leftrightarrow\frac{t+1+2t}{t\left(t+1\right)}=\frac{6}{t+2}$ $\Leftrightarrow\frac{3t+1}{t^2+t}=\frac{6}{t+2}$

$\Leftrightarrow\left(3t+1\right)\left(t+2\right)=6\left(t^2+t\right)$

$\Leftrightarrow3t^2+7t+2=6t^2+6t$

$\Leftrightarrow3t^2-t-2=0$

$\Leftrightarrow3t^2-3t+2t-2=0$

$\Leftrightarrow\left(3t+2\right)\left(t-1\right)=0$

$\Leftrightarrow t-1=0$ ( do $3t+2=3x^2-9x+11$$=3\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{17}{12}\right)=3\left[\left(x-\frac{3}{2}\right)^2+\frac{17}{12}\right]>0\forall x$)

$\Leftrightarrow x^2-3x+3=1$

$\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=1$

$\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}$

$\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{2}=\frac{1}{2}\x-\frac{3}{2}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=1\end{matrix}\right.\left(TM\right)$

Vậy tập nghiệm của pt đã cho là $S=\left\{1;2\right\}$Câu trả lời: + Đặt $t=x^2-3x+3$ thì pt đã cho trở thành :

$\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}$

$\Leftrightarrow\frac{t+1+2t}{t\left(t+1\right)}=\frac{6}{t+2}$ $\Leftrightarrow\frac{3t+1}{t^2+t}=\frac{6}{t+2}$

$\Leftrightarrow\left(3t+1\right)\left(t+2\right)=6\left(t^2+t\right)$

$\Leftrightarrow3t^2+7t+2=6t^2+6t$

$\Leftrightarrow3t^2-t-2=0$

$\Leftrightarrow3t^2-3t+2t-2=0$

$\Leftrightarrow\left(3t+2\right)\left(t-1\right)=0$

$\Leftrightarrow t-1=0$ ( do $3t+2=3x^2-9x+11$$=3\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{17}{12}\right)=3\left[\left(x-\frac{3}{2}\right)^2+\frac{17}{12}\right]>0\forall x$)

$\Leftrightarrow x^2-3x+3=1$

$\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=1$

$\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}$

$\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{2}=\frac{1}{2}\x-\frac{3}{2}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=1\end{matrix}\right.\left(TM\right)$

Vậy tập nghiệm của pt đã cho là $S=\left\{1;2\right\}$

Nguyễn Việt Lâm · Answer

$\frac{1}{x^2-3x+3}-1+\frac{2}{x^2-3x+4}-1+2-\frac{6}{x^2-3x+5}=0$

$\Leftrightarrow\frac{-x^2+3x-2}{x^2-3x+3}+\frac{-x^2+3x-2}{x^2-3x+4}-\frac{2\left(-x^2+3x-2\right)}{x^2-3x+5}=0$

$\Leftrightarrow\left(-x^2+3x-2\right)\left(\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}-x^2+3x-2=0\left(1\right)\\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}=0\left(2\right)\end{matrix}\right.$

$\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=2\x=1\end{matrix}\right.$

$\left(2\right)\Leftrightarrow\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}=0$

Do $\left\{{}\begin{matrix}\frac{1}{x^2-3x+3}>\frac{1}{x^2-3x+5}\\frac{1}{x^2-3x+4}>\frac{1}{x^2-3x+5}\end{matrix}\right.$ $\forall x\Rightarrow\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}>0$

$\Rightarrow\left(2\right)$ vô nghiệmCâu trả lời: $\frac{1}{x^2-3x+3}-1+\frac{2}{x^2-3x+4}-1+2-\frac{6}{x^2-3x+5}=0$