Có bao nhiêu số nguyên $x$sao cho ứng với mỗi $x$ tồn tại $y \in \left[ 2 ; 8 \left]\right.$ thoả mãn $\left(\right. y - x \right) \left(log\right)_{2} \left( x + y \right) = y + x^{2}$

Có bao nhiêu số nguyên $x$sao cho ứng với mỗi $x$ tồn tại $y \in \left[ 2 ; 8 \left]\right.$ thoả mãn $\left(\right. y - x \right) \left(log\right)_{2} \left( x + y \right) = y + x^{2}$
A. $5$. B. $8$. C. $4$. D. $7$.
Lời giải
$\left( y - x \right) \left(log\right)_{2} \left( x + y \right) = y + x^{2} \left( 1 \right) .$
Do $y \in \left[\right. 2 ; 8 \left]\right. \Rightarrow 2 \leq y + x^{2} \leq 8 \Rightarrow V P \left( 1 \right) > 0 .$
+ Nếu $\left{\right. 0 < x + y \leq 1 \\ y - x \geq 0 \\ y \in \left[\right. 2 ; 8 \left]\right. \Leftrightarrow \left{\right. 0 < x + y \leq 1 \\ y \geq x \\ y \in \left[\right. 2 ; 8 \left]\right. \Rightarrow \left{\right. V T \left( 1 \right) \leq 0 \\ V T \left( 1 \right) \geq 2$ không có giá trị nào của $x$thỏa mãn.
+ Nếu $\left{\right. 0 < x + y \leq 1 \\ y - x < 0 \\ y \in \left[\right. 2 ; 8 \left]\right. \Leftrightarrow \left{\right. 0 < x + y \leq 1 \\ y < x \\ y \in \left[\right. 2 ; 8 \left]\right. \Rightarrow \left{\right. 2 y \leq 1 \\ y \in \left[\right. 2 ; 8 \left]\right.$ không thỏa mãn.
+ Nếu $\left{\right. 1 < x + y \leq 2 \\ y - x \geq 0 \\ y \in \left[\right. 2 ; 8 \left]\right. \Leftrightarrow \left{\right. 1 < x + y \leq 2 \\ y \geq x \\ y \in \left[\right. 2 ; 8 \left]\right. \Rightarrow \left( y - x \right) \left(log\right)_{2} \left( x + y \right) \leq y - x < y + x^{2}$ không có giá trị nào của $x$thỏa mãn.
+ Nếu $\left{ x + y > 2 \\ y - x > 0 \\ y \in \left[\right. 2 ; 8 \left]\right. .$
Đặt $f \left(\right. y \right) = \left( y - x \right) \left(log\right)_{2} \left( x + y \right) - y - x^{2} , y \in \left[ 2 ; 4 \left]\right. .$
$f^{'} \left(\right. y \right) = \left(log\right)_{2} \left( x + y \right) + \dfrac{\left( y - x \right)}{\left( x + y \right) ln2} - 1 > 0$
Do $\left{\right. x + y > 2 \\ y \in \left[\right. 2 ; 8 \left]\right. \Rightarrow \left(log\right)_{2} \left( x + y \right) > 1 \Rightarrow \left(og\right)_{2} \left( x + y \right) - 1 + \dfrac{\left( y - x \right)}{\left( x + y \right) ln2} > 0 .$
Do hàm số đồng biến. để phuong trình $f \left( y \right) = 0$có nghiệm khi
$\left{\right. f \left( 2 \right) \leq 0 \\ f \left( 8 \right) \geq 0 \\ x < y \\ y \in \left[ 2 ; 8 \left]\right. \Leftrightarrow \left{\right. \left(\right. 2 - x \right) \left(log\right)_{2} \left( x + 2 \right) - 2 - x^{2} \leq 0 \\ \left( 8 - x \right) \left(log\right)_{2} \left( x + 8 \right) - 8 - x^{2} \geq 0 \\ x < y \\ y \in \left[ 2 ; 8 \left]\right. \Leftrightarrow \left{\right. \left(\right. 2 - x \right) \left(log\right)_{2} \left( x + 2 \right) \leq 2 + x^{2} \\ \left( 8 - x \right) \left(log\right)_{2} \left( x + 8 \right) \geq 2 + x^{2} \\ - 2 < x < y , x \in \mathbb{Z} \\ y \in \left[ 2 ; 8 \left]\right. .$
$\left{\right. \left(\right. 2 - x \right) \left(log\right)_{2} \left( x + 2 \right) \leq 2 + x^{2} \\ \left( 8 - x \right) \left(log\right)_{2} \left( x + 8 \right) \geq 2 + x^{2} \\ x \in \left{ - 1 , 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 \right} \Rightarrow x = - 1 , x = 0 , x = 1 , x = 2 , x = 3 .$
Vậy có 5 giá trị nguyên thỏa mãn.