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Câu hỏi: Gọi ${{m}_{0}}$ là giá trị nhỏ nhất để bất phương trình
$1+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le -{{\log }_{2}}\left( x+1 \right)$ có nghiệm.
Chọn đáp án đúng trong các khẳng định sau:
A. ${{m}_{0}}\in \left( -9;-8 \right)$
B. ${{m}_{0}}\in \left( 8;9 \right)$
C. ${{m}_{0}}\in \left( 9;10 \right)$
D. ${{m}_{0}}\in (-10;9)$
$1+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le -{{\log }_{2}}\left( x+1 \right)$ có nghiệm.
Chọn đáp án đúng trong các khẳng định sau:
A. ${{m}_{0}}\in \left( -9;-8 \right)$
B. ${{m}_{0}}\in \left( 8;9 \right)$
C. ${{m}_{0}}\in \left( 9;10 \right)$
D. ${{m}_{0}}\in (-10;9)$
Cách giải:
$1+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le -{{\log }_{2}}\left( x+1 \right)\left( -1<x<2 \right)$
$\Leftrightarrow $ $1+{{\log }_{2}}\left( x+1 \right)+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow {{\log }_{2}}\left( 2x+2 \right)+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow 2{{\log }_{2}}\sqrt{2x+2}+2{{\log }_{2}}\sqrt{2-x}-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow {{\log }_{2}}\sqrt{2x+2}+{{\log }_{2}}\sqrt{2-x}-{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow {{\log }_{2}}\dfrac{\sqrt{2x+2}\sqrt{2-x}}{m-\dfrac{x}{2}+4\left( \sqrt{2x-x}+\sqrt{2x+2} \right)}\le 0$
$\Leftrightarrow \dfrac{\sqrt{2x+2}\sqrt{2-x}}{m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right)}\le 1$
$\Leftrightarrow \sqrt{2x+2}\sqrt{2-x}\le m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right)$
$\Leftrightarrow 2\sqrt{2x+2}\sqrt{2-x}\le 2m-x+8\left( \sqrt{2-x}+\sqrt{2x+2} \right)$
$\Leftrightarrow 2\sqrt{2x+2}\sqrt{2-x}+x-8\left( \sqrt{2-x}+\sqrt{2x+2} \right)\le 2m$
Đặt $t=\sqrt{2-x}+\sqrt{2x+2}$ ta có
$t'\left( x \right)=\dfrac{-1}{2\sqrt{2-x}}+\dfrac{2}{2\sqrt{2x+2}}=0$
$\Leftrightarrow \dfrac{2}{2\sqrt{2x+2}}=\dfrac{1}{2\sqrt{2-x}}$
$\begin{aligned}
& \Leftrightarrow 2\sqrt{2-x}=\sqrt{2x+2} \\
& \Leftrightarrow 4\left( 2-x \right)=2x+2 \\
& \Leftrightarrow 8-4x=2x+2 \\
& \Leftrightarrow x=1\left( tm \right) \\
\end{aligned}$
$t\left( -1 \right)=\sqrt{3},t\left( 2 \right)=\sqrt{6},t\left( 1 \right)=3$ Do đó $t\in \left( \sqrt{3};3 \right)$
Ta có:
${{t}^{2}}=2-x+2x+2+2\sqrt{2-x}\sqrt{2x+2}$
${{t}^{2}}=4+x+2\sqrt{2-x}\sqrt{2x+2}$
$\Leftrightarrow x+2\sqrt{2-x}\sqrt{2x+2}={{t}^{2}}-4$
Bất phương trình trở thành $g\left( t \right)={{t}^{2}}-4-8t\le 2m\left( * \right)$ với $t\in \left( \sqrt{3};3 \right).~$
Để bất phương trình ban đầu có nghiệm thì (*) có nghiệm $t\in \left( \sqrt{3};3 \right)\Rightarrow 2m\ge \underset{\left[ \sqrt{3};3 \right]}{\mathop{min}} g\left( t \right)$.
Ta có: $g'\left( t \right)=2t-8=0\Leftrightarrow t=4\left( ktm \right)$.
$g\left( \sqrt{3} \right)=-1-8\sqrt{3},g\left( 3 \right)=-19$
$\underset{\left[ \sqrt{3};3 \right]}{\mathop{\min }} g\left( t \right)=-19\Leftrightarrow 2m\ge -19\Leftrightarrow m\ge -9,5$
Vậy ${{m}_{0}}=-9,5\in \left( -10;-9 \right).~$
$1+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le -{{\log }_{2}}\left( x+1 \right)\left( -1<x<2 \right)$
$\Leftrightarrow $ $1+{{\log }_{2}}\left( x+1 \right)+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow {{\log }_{2}}\left( 2x+2 \right)+{{\log }_{2}}\left( 2-x \right)-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow 2{{\log }_{2}}\sqrt{2x+2}+2{{\log }_{2}}\sqrt{2-x}-2{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow {{\log }_{2}}\sqrt{2x+2}+{{\log }_{2}}\sqrt{2-x}-{{\log }_{2}}\left( m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right) \right)\le 0$
$\Leftrightarrow {{\log }_{2}}\dfrac{\sqrt{2x+2}\sqrt{2-x}}{m-\dfrac{x}{2}+4\left( \sqrt{2x-x}+\sqrt{2x+2} \right)}\le 0$
$\Leftrightarrow \dfrac{\sqrt{2x+2}\sqrt{2-x}}{m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right)}\le 1$
$\Leftrightarrow \sqrt{2x+2}\sqrt{2-x}\le m-\dfrac{x}{2}+4\left( \sqrt{2-x}+\sqrt{2x+2} \right)$
$\Leftrightarrow 2\sqrt{2x+2}\sqrt{2-x}\le 2m-x+8\left( \sqrt{2-x}+\sqrt{2x+2} \right)$
$\Leftrightarrow 2\sqrt{2x+2}\sqrt{2-x}+x-8\left( \sqrt{2-x}+\sqrt{2x+2} \right)\le 2m$
Đặt $t=\sqrt{2-x}+\sqrt{2x+2}$ ta có
$t'\left( x \right)=\dfrac{-1}{2\sqrt{2-x}}+\dfrac{2}{2\sqrt{2x+2}}=0$
$\Leftrightarrow \dfrac{2}{2\sqrt{2x+2}}=\dfrac{1}{2\sqrt{2-x}}$
$\begin{aligned}
& \Leftrightarrow 2\sqrt{2-x}=\sqrt{2x+2} \\
& \Leftrightarrow 4\left( 2-x \right)=2x+2 \\
& \Leftrightarrow 8-4x=2x+2 \\
& \Leftrightarrow x=1\left( tm \right) \\
\end{aligned}$
$t\left( -1 \right)=\sqrt{3},t\left( 2 \right)=\sqrt{6},t\left( 1 \right)=3$ Do đó $t\in \left( \sqrt{3};3 \right)$
Ta có:
${{t}^{2}}=2-x+2x+2+2\sqrt{2-x}\sqrt{2x+2}$
${{t}^{2}}=4+x+2\sqrt{2-x}\sqrt{2x+2}$
$\Leftrightarrow x+2\sqrt{2-x}\sqrt{2x+2}={{t}^{2}}-4$
Bất phương trình trở thành $g\left( t \right)={{t}^{2}}-4-8t\le 2m\left( * \right)$ với $t\in \left( \sqrt{3};3 \right).~$
Để bất phương trình ban đầu có nghiệm thì (*) có nghiệm $t\in \left( \sqrt{3};3 \right)\Rightarrow 2m\ge \underset{\left[ \sqrt{3};3 \right]}{\mathop{min}} g\left( t \right)$.
Ta có: $g'\left( t \right)=2t-8=0\Leftrightarrow t=4\left( ktm \right)$.
$g\left( \sqrt{3} \right)=-1-8\sqrt{3},g\left( 3 \right)=-19$
$\underset{\left[ \sqrt{3};3 \right]}{\mathop{\min }} g\left( t \right)=-19\Leftrightarrow 2m\ge -19\Leftrightarrow m\ge -9,5$
Vậy ${{m}_{0}}=-9,5\in \left( -10;-9 \right).~$
Đáp án D.