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Câu hỏi: Cho các số thực $x,y,z$ thỏa mãn ${{3}^{x}}={{5}^{y}}={{15}^{\dfrac{2023}{x+y}-z}}$. Tính giá trị biểu thức $S=xy+yz+zx$ bằng
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A. $2022.$
B. $1011.$
C. $2023.$
D. $1012.$
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A. $2022.$
B. $1011.$
C. $2023.$
D. $1012.$
Điều kiện $x+y\ne 0$.
Đặt ${{3}^{x}}={{5}^{y}}={{15}^{\dfrac{2023}{x+y}-z}}=t\Rightarrow \left\{ \begin{aligned}
& x={{\log }_{3}}t \\
& y={{\log }_{5}}t \\
& \dfrac{2023}{x+y}-z={{\log }_{15}}t \\
\end{aligned} \right.$
Mà ${{\log }_{15}}t=\dfrac{1}{{{\log }_{t}}15}=\dfrac{1}{{{\log }_{t}}3+{{\log }_{t}}5}=\dfrac{{{\log }_{3}}t.{{\log }_{5}}t}{{{\log }_{3}}t+{{\log }_{5}}t}$
$\Rightarrow \dfrac{2023}{x+y}-z=\dfrac{xy}{x+y}$
$\Rightarrow 2023-z\left( x+y \right)=xy$
$\Leftrightarrow xy+yz+zx=2023$.
Đặt ${{3}^{x}}={{5}^{y}}={{15}^{\dfrac{2023}{x+y}-z}}=t\Rightarrow \left\{ \begin{aligned}
& x={{\log }_{3}}t \\
& y={{\log }_{5}}t \\
& \dfrac{2023}{x+y}-z={{\log }_{15}}t \\
\end{aligned} \right.$
Mà ${{\log }_{15}}t=\dfrac{1}{{{\log }_{t}}15}=\dfrac{1}{{{\log }_{t}}3+{{\log }_{t}}5}=\dfrac{{{\log }_{3}}t.{{\log }_{5}}t}{{{\log }_{3}}t+{{\log }_{5}}t}$
$\Rightarrow \dfrac{2023}{x+y}-z=\dfrac{xy}{x+y}$
$\Rightarrow 2023-z\left( x+y \right)=xy$
$\Leftrightarrow xy+yz+zx=2023$.
Đáp án C.
