Giải thích các bước giải:
Bài 1:
$P=\dfrac{x+2}{x-1}(ĐKXĐ: x\neq 1)$
$Q=\dfrac{x-1}{x}+\dfrac{2x+1}{x^{2}+x} (ĐKXĐ: x\neq 0;x\neq -1)$
a) Ta có: $\left | x-2 \right |=1$
$\Rightarrow \left[ \begin{array}{l}x-2=1\\x-2=-1\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}x=3\\x=1\end{array} \right.$
* Với $x=3$, giá trị biểu thức P là:
$\dfrac{3+2}{3-1}=\dfrac{5}{2}=2,5$
* Với $x=1$ không thỏa mãn điều kiện xác định
b) $Q=\dfrac{x-1}{x}+\dfrac{2x+1}{x^{2}+x}$
$Q=\dfrac{x-1}{x}+\dfrac{2x+1}{x(x+1)}$
$Q=\dfrac{(x-1)(x+1)}{x(x+1)}+\dfrac{2x+1}{x(x+1)}$
$Q=\dfrac{x^{2}-1+2x+1}{x(x+1)}$
$Q=\dfrac{x^{2}+2x}{x(x+1)}$
$Q=\dfrac{x(x+2)}{x(x+1)}$
$Q=\dfrac{x+2}{x+1}$
c) $M=\dfrac{P}{Q}=\dfrac{x+2}{x-1}:\dfrac{x+2}{x+1}$
$M=\dfrac{P}{Q}=\dfrac{x+2}{x-1}.\dfrac{x+1}{x+2}$
$M=\dfrac{P}{Q}=\dfrac{x+1}{x-1}$
Ta có:
$\Rightarrow M=\dfrac{x+1}{x-1}$
$\Leftrightarrow M=\dfrac{(x-1)+2}{x-1}$
$\Leftrightarrow 1+\dfrac{2}{x-1}$
Để $M\in \mathbb{Z}\Rightarrow (x-1)\in Ư(2)$
mà $Ư(2)=\left \{ ±1;±2 \right \}$
$\Rightarrow x\in \left \{ -1;0;2;3 \right \}$
Bài 2:
a) $x^{3}+5x^{2}-3x=15$
$\Leftrightarrow x^{3}+5x^{2}-3x-15=0$
$\Leftrightarrow x^{2}(x+5)-3(x+5)=0$
$\Leftrightarrow (x+5)(x^{2}-3)=0$
$\Leftrightarrow (x+5)(x-\sqrt{3})(x+\sqrt{3})=0$
$\Leftrightarrow \left[ \begin{array}{l}x+5=0\\x-\sqrt{3}=0\\x+\sqrt{3}=0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x=-5\\x=\sqrt{3}\\x=-\sqrt{3}\end{array} \right.$
Vậy $x\in \left \{ -5;\sqrt{3};-\sqrt{3} \right \}$
b) $\dfrac{x^{2}-2}{2x-3}=2$
$\Leftrightarrow x^{2}-2=2(2x-3)$
$\Leftrightarrow x^{2}-2=4x-6$
$\Leftrightarrow x^{2}-4x-2+6=0$
$\Leftrightarrow x^{2}-4x+4=0$
$\Leftrightarrow (x-2)^{2}=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
Vậy $x=2$
c) $(3x-1)(2x+7)+(x+1)(5-6x)=16$
$\Leftrightarrow 6x^{2}+21x-2x-7+5x-6x^{2}+5-6x-16=0$
$\Leftrightarrow 18x-18=0$
$\Leftrightarrow 18(x-1)=0$
$\Leftrightarrow x-1=0$
$\Leftrightarrow x=1$
Vậy $x=1$
