$Bài 1:$
$c)\frac{3x-2}{2x-1}+\frac{1-x}{2x-1}=\frac{3x-2+1-x}{2x-1}=\frac{2x-1}{2x-1}=1$
$Bài 3:$
$a) P=\frac{2-x}{x^3+1}+\frac{x-1}{x+1}:\frac{x^2-x+1}{x+1}$
$P=\frac{2-x}{(x+1)(x^2-x+1)}+\frac{(x-1)(x+1)}{(x+1)(x^2-x+1)}$
$P=\frac{2-x+(x-1)(x+1)}{(x+1)(x^2-x+1)}$
$P=\frac{2-x+x-1}{x^2-x+1}=\frac{1}{x^2-x+1}$
b) Theo đề ta có: $P=\frac{1}{P}$
$⇔(\frac{1}{x^2-x+1})^2=1$
$⇔(x^2-x+1)^2=1$
$⇔x^4-x^2+1=1$
$⇔x^2(x^2+1)=0$
$⇔x^2(x+1)(x-1)=0$
$⇔x=0$ $hoặc$
⇔\(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
$Vậy$ $để$ $P=\frac{1}{P}$ $thì$
$x=0$ $hoặc$ $x=1$ $hoặc$ $x=-1$
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