Question

Giúp mik vs

Answer 1

Lời giải:
a. ĐKXĐ: $x\neq 0; x\neq \pm 1$

\(P=\left[\frac{2}{3x}-\frac{2}{x+1}(\frac{x+1}{3x}-(x+1))\right].\frac{x}{x-1}\)

\(=(\frac{2}{3x}-\frac{2}{3x}+2).\frac{x}{x-1}=\frac{2x}{x-1}\)

b. Để $P$ nguyên thì $\frac{2x}{x-1}$ nguyên 

Với $x$ nguyên thì điều này xảy ra khi $2x\vdots x-1$
$\Leftrightarrow 2(x-1)+2\vdots x-1$

$\Leftrightarrow 2\vdots x-1$

$\Rightarrow x-1\in\left\{\pm 1; \pm 2\right\}$

$\Rightarrow x\in \left\{2; 0; 3; -1\right\}$

c. Để $P\leq 1\Leftrightarrow \frac{2x}{x-1}-1\leq 0$

$\Leftrightarrow \frac{x+1}{x-1}\leq 0$

$\Leftrightarrow -1\leq x< 1$

Kết hợp đkxđ suy ra $-1< x< 1$ và $x\neq 0$

Answer 2

a: \(P=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x\left(x+1\right)}{3x}\right):\dfrac{x-1}{x}\)

\(=\left(\dfrac{2}{3x}-\dfrac{2\left(x+1\right)\left(1-3x\right)}{3x\left(x+1\right)}\right)\cdot\dfrac{x}{x-1}\)

\(=\left(\dfrac{2}{3x}+\dfrac{2\left(3x-1\right)}{3x}\right)\cdot\dfrac{x}{x-1}\)

\(=\dfrac{2+6x-2}{3x}\cdot\dfrac{x}{x-1}=\dfrac{6x}{3}\cdot\dfrac{1}{x-1}=\dfrac{2}{x-1}\)

b: Để P nguyên thì \(x-1\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{2;3\right\}\)

c: Để P<=1 thì P-1<=0

\(\Leftrightarrow\dfrac{2-x+1}{x-1}< =0\)

=>(x-2)/(x-1)>=0

\(\Leftrightarrow\left[{}\begin{matrix}x>=2\\0\ne x< 1\end{matrix}\right.\)