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Câu hỏi: Có bao nhiêu cặp số $\left( x;y \right)$ thoả mãn
${{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-{{2}^{2+{{\log }_{3}}x{{\log }_{2}}y}}+8 \right)={{\log }_{3}}\left[ 7-\left( {{x}^{2}}+{{y}^{3}}-2025 \right)\sqrt{{{x}^{2}}+{{y}^{3}}-2022} \right]$ ?
A. $2$.
B. $1$.
C. $3$.
D. $0$
${{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-{{2}^{2+{{\log }_{3}}x{{\log }_{2}}y}}+8 \right)={{\log }_{3}}\left[ 7-\left( {{x}^{2}}+{{y}^{3}}-2025 \right)\sqrt{{{x}^{2}}+{{y}^{3}}-2022} \right]$ ?
A. $2$.
B. $1$.
C. $3$.
D. $0$
Ta có:
$\begin{aligned}
& {{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-{{2}^{2+{{\log }_{3}}x{{\log }_{2}}y}}+8 \right)={{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-4.{{\left( {{2}^{{{\log }_{2}}y}} \right)}^{{{\log }_{3}}x}}+8 \right) \\
& ={{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-4.{{y}^{{{\log }_{3}}x}}+4+4 \right)={{\log }_{2}}\left( {{\left( {{y}^{2{{\log }_{3}}x}}-2 \right)}^{2}}+4 \right)\ge {{\log }_{2}}4=2 \\
\end{aligned}$
${{\log }_{3}}\left[ 7-\left( {{x}^{2}}+{{y}^{3}}-2025 \right)\sqrt{{{x}^{2}}+{{y}^{3}}-2022} \right]={{\log }_{3}}\left[ 7-\left( {{x}^{2}}+{{y}^{3}}-2022 \right)\sqrt{{{x}^{2}}+{{y}^{3}}-2022}+3\sqrt{{{x}^{2}}+{{y}^{3}}-2022} \right]$
Đặt $t=\sqrt{{{x}^{2}}+{{y}^{3}}-2022},t\ge 0$
Xét hàm $-{{t}^{3}}+3t+7$ trên $\left[ 0;+\infty \right)$ thì hàm này có bảng biến thiên như sau:
Vậy $\underset{t\ge 0}{\mathop{\max }} \left( -{{t}^{3}}+3t+7 \right)=y\left( 1 \right)=9$
$\Rightarrow VP\le {{\log }_{3}}9=2$
Dấu $''=''$ xảy ra $\Leftrightarrow \left\{ \begin{matrix}
\sqrt{{{x}^{2}}+{{y}^{3}}-2022}=1 \\
{{y}^{2{{\log }_{3}}x}}-2=0 \\
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix}
{{x}^{2}}+{{y}^{3}}=2023 \\
{{y}^{2{{\log }_{3}}x}}=2 \\
\end{matrix} \right.$
$\begin{aligned}
& \Rightarrow \underbrace{{{x}^{2}}+{{8}^{\dfrac{\ln 8}{\ln x}}}}_{g\left( x \right)}=2023 \\
& g'\left( x \right)=2x-{{8}^{\dfrac{\ln 3}{\ln 8}}}.\dfrac{\ln 3}{x.{{\ln }^{2}}x}.\ln 8 \\
& g'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x\approx 0,34 \\ x\approx 2,91 \\
\end{matrix} \right. \\
\end{aligned}$
$\Rightarrow g\left( x \right)=2023$ có 2 nghiệm.
Vậy có 2 cặp $\left( x,y \right)$ thoả mãn.
$\begin{aligned}
& {{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-{{2}^{2+{{\log }_{3}}x{{\log }_{2}}y}}+8 \right)={{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-4.{{\left( {{2}^{{{\log }_{2}}y}} \right)}^{{{\log }_{3}}x}}+8 \right) \\
& ={{\log }_{2}}\left( {{y}^{2{{\log }_{3}}x}}-4.{{y}^{{{\log }_{3}}x}}+4+4 \right)={{\log }_{2}}\left( {{\left( {{y}^{2{{\log }_{3}}x}}-2 \right)}^{2}}+4 \right)\ge {{\log }_{2}}4=2 \\
\end{aligned}$
${{\log }_{3}}\left[ 7-\left( {{x}^{2}}+{{y}^{3}}-2025 \right)\sqrt{{{x}^{2}}+{{y}^{3}}-2022} \right]={{\log }_{3}}\left[ 7-\left( {{x}^{2}}+{{y}^{3}}-2022 \right)\sqrt{{{x}^{2}}+{{y}^{3}}-2022}+3\sqrt{{{x}^{2}}+{{y}^{3}}-2022} \right]$
Đặt $t=\sqrt{{{x}^{2}}+{{y}^{3}}-2022},t\ge 0$
Xét hàm $-{{t}^{3}}+3t+7$ trên $\left[ 0;+\infty \right)$ thì hàm này có bảng biến thiên như sau:
$\Rightarrow VP\le {{\log }_{3}}9=2$
Dấu $''=''$ xảy ra $\Leftrightarrow \left\{ \begin{matrix}
\sqrt{{{x}^{2}}+{{y}^{3}}-2022}=1 \\
{{y}^{2{{\log }_{3}}x}}-2=0 \\
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix}
{{x}^{2}}+{{y}^{3}}=2023 \\
{{y}^{2{{\log }_{3}}x}}=2 \\
\end{matrix} \right.$
$\begin{aligned}
& \Rightarrow \underbrace{{{x}^{2}}+{{8}^{\dfrac{\ln 8}{\ln x}}}}_{g\left( x \right)}=2023 \\
& g'\left( x \right)=2x-{{8}^{\dfrac{\ln 3}{\ln 8}}}.\dfrac{\ln 3}{x.{{\ln }^{2}}x}.\ln 8 \\
& g'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x\approx 0,34 \\ x\approx 2,91 \\
\end{matrix} \right. \\
\end{aligned}$
Vậy có 2 cặp $\left( x,y \right)$ thoả mãn.
Đáp án A.