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Câu hỏi: Có bao nhiêu cặp số nguyên $\left( x; y \right); y\in \left[ 0; {{2023}^{3}} \right]$ thỏa mãn phương trình ${{\log }_{4}}\left( x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)={{\log }_{2}}\left( y-x \right)$ ?
A. $90854$.
B. $90990$.
C. ${{2021}^{2}}$.
D. ${{2021}^{2}}-1$.
A. $90854$.
B. $90990$.
C. ${{2021}^{2}}$.
D. ${{2021}^{2}}-1$.
Điều kiện:$\left\{ \begin{aligned}
& y-x>0 \\
& x\ge -\dfrac{1}{4} \\
& x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}>0 \\
\end{aligned} \right.$.
Ta có: ${{\log }_{4}}\left( x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)={{\log }_{2}}\left( y-x \right)$ $\Leftrightarrow {{\log }_{4}}{{\left( \dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)}^{2}}={{\log }_{2}}\left( y-x \right)$
$\Leftrightarrow {{\log }_{2}}\left( \dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)={{\log }_{2}}\left( y-x \right)\Leftrightarrow $ $y-x=\left( \dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)\Leftrightarrow y={{\left( \sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2} \right)}^{2}}$.
Vì $y\in \mathbb{Z}\Rightarrow \sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\in \mathbb{Z}\Rightarrow \sqrt{x+\dfrac{1}{4}}=\dfrac{2m+1}{2}\left( m\in \mathbb{N} \right)\Rightarrow x={{m}^{2}}+m$ ; khi đó $y={{\left( m+1 \right)}^{2}}$.
Mà $y\in \left[ 0; {{2023}^{3}} \right]\Rightarrow {{\left( m+1 \right)}^{2}}\le {{2023}^{3}}\Rightarrow 0\le m\le 2023\sqrt{2023}-1\approx 90989,03$.
Do đó có $90990$ giá trị của $m$, ứng với đó có $90990$ cặp $\left( x; y \right)$ thỏa mãn yêu cầu bài toán.
& y-x>0 \\
& x\ge -\dfrac{1}{4} \\
& x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}>0 \\
\end{aligned} \right.$.
Ta có: ${{\log }_{4}}\left( x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)={{\log }_{2}}\left( y-x \right)$ $\Leftrightarrow {{\log }_{4}}{{\left( \dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)}^{2}}={{\log }_{2}}\left( y-x \right)$
$\Leftrightarrow {{\log }_{2}}\left( \dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)={{\log }_{2}}\left( y-x \right)\Leftrightarrow $ $y-x=\left( \dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}} \right)\Leftrightarrow y={{\left( \sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2} \right)}^{2}}$.
Vì $y\in \mathbb{Z}\Rightarrow \sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\in \mathbb{Z}\Rightarrow \sqrt{x+\dfrac{1}{4}}=\dfrac{2m+1}{2}\left( m\in \mathbb{N} \right)\Rightarrow x={{m}^{2}}+m$ ; khi đó $y={{\left( m+1 \right)}^{2}}$.
Mà $y\in \left[ 0; {{2023}^{3}} \right]\Rightarrow {{\left( m+1 \right)}^{2}}\le {{2023}^{3}}\Rightarrow 0\le m\le 2023\sqrt{2023}-1\approx 90989,03$.
Do đó có $90990$ giá trị của $m$, ứng với đó có $90990$ cặp $\left( x; y \right)$ thỏa mãn yêu cầu bài toán.
Đáp án B.