C = (x^2 + x / x^3 + x^2 +x +1+ 1/ x^2 +1 ) : ( 1/x-1 – 2x/x^3 -x^2 +x -1)
a) tìm đk xác định
b) RG
c) tìm x nguyên để C nguyên
C = (x^2 + x / x^3 + x^2 +x +1+ 1/ x^2 +1 ) : ( 1/x-1 – 2x/x^3 -x^2 +x -1)
a) tìm đk xác định
b) RG
c) tìm x nguyên để C nguyên
Đáp án:
$\begin{array}{l}
C = \left( {\frac{{{x^2} + x}}{{{x^3} + {x^2} + x + 1}} + \frac{1}{{{x^2} + 1}}} \right):\left( {\frac{1}{{x – 1}} – \frac{{2x}}{{{x^3} – {x^2} + x – 1}}} \right)\\
a)Đkxđ:\\
\left\{ \begin{array}{l}
{x^3} + {x^2} + x + 1 \ne 0\\
{x^2} + 1 \ne 0\left( {ld} \right)\\
x – 1 \ne 0\\
{x^3} – {x^2} + x – 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {{x^2} + 1} \right) \ne 0\\
\left( {x – 1} \right)\left( {{x^2} + 1} \right) \ne 0\\
x \ne 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ne – 1\\
x \ne 1
\end{array} \right.\\
b)Đkxđ:x \ne 1;x \ne – 1\\
C = \left( {\frac{{{x^2} + x}}{{{x^3} + {x^2} + x + 1}} + \frac{1}{{{x^2} + 1}}} \right):\left( {\frac{1}{{x – 1}} – \frac{{2x}}{{{x^3} – {x^2} + x – 1}}} \right)\\
= \left( {\frac{{x\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{1}{{{x^2} + 1}}} \right):\left( {\frac{1}{{x – 1}} – \frac{{2x}}{{\left( {{x^2} + 1} \right)\left( {x – 1} \right)}}} \right)\\
= \left( {\frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right):\frac{{{x^2} + 1 – 2x}}{{\left( {{x^2} + 1} \right)\left( {x – 1} \right)}}\\
= \frac{{x + 1}}{{{x^2} + 1}}.\frac{{\left( {{x^2} + 1} \right)\left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}\\
= \frac{{x + 1}}{{x – 1}}\\
c)Đkxđ:x \ne 1;x \ne – 1\\
C = \frac{{x + 1}}{{x – 1}} = \frac{{x – 1 + 2}}{{x – 1}} = 1 + \frac{2}{{x – 1}}\\
C \in Z \Rightarrow \frac{2}{{x – 1}} \in Z\\
\Rightarrow 2 \vdots \left( {x – 1} \right)\\
\Rightarrow \left( {x – 1} \right) \in {\rm{\{ }} – 2; – 1;1;2\} \\
\Rightarrow x \in {\rm{\{ }} – 1;0;2;3\} \\
Mà:x \ne 1;x \ne – 1\\
\Rightarrow x \in {\rm{\{ }}0;2;3\}
\end{array}$
a, Để C xác định
$\Leftrightarrow x\neq\pm1$
b, $C=(\dfrac{x^2+x}{x^3+x^2+x+1}+\dfrac{1}{x^2+1}):(\dfrac{1}{x-1}-\dfrac{2x}{x^3-x^2+x-1})$ $(x\neq\pm1)$
$C=(\dfrac{x^2+x}{x^2(x+1)+(x+1)}+\dfrac{1}{x^2+1}):(\dfrac{1}{x-1}-\dfrac{2x}{x^2(x-1)+(x-1)})$
$C=\dfrac{x(x+1)}{(x+1)(x^2+1)}+\dfrac{1}{x^2+1}):(\dfrac{1}{x-1}-\dfrac{2x}{(x^2+1)(x-1)})$
$C=\dfrac{x+1}{x^2+1}:\dfrac{x^2+1-2x}{(x-1)(x^2+1)}$
$C=\dfrac{x+1}{x^2+1}\cdot\dfrac{(x-1)(x^2+1)}{(x-1)^2}$
$C=\dfrac{x+1}{x-1}$
Vậy $C=\dfrac{x+1}{x-1}$ với $x\neq\pm1$
c, $C=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}$
Để C nguyên
$\dfrac{2}{x-1}$ nguyên
`⇒x-1∈Ư(2)={±1;±2}`.
· $x-1=1⇒x=2(tm)$
· $x-1=-1⇒x=0(ktm)$
· $x-1=2⇒x=3(tm)$
· $x-1=-2⇒x=-1(ktm)$
Vậy để C nguyên thì `x\in{2;3}`.