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Câu hỏi: Cho ${x,y}$ là hai số nguyên thỏa mãn: ${{3^x}{.6^y} = \dfrac{{{2^{15}}{{.6}^{40}}}}{{{9^{50}}{{.12}^{25}}}}}$. Tính ${x.y}$ ?
A. ${ - 755}$.
B. ${ - 450}$.
C. ${ - 425}$.
D. ${ - 445}$.
A. ${ - 755}$.
B. ${ - 450}$.
C. ${ - 425}$.
D. ${ - 445}$.
Ta có:
$VT={{3}^{x}}{{.6}^{y}}={{3}^{x}}.{{\left( 2.3 \right)}^{y}}={{2}^{y}}.3{{x}^{x+y}}$
$VP=\dfrac{{{2}^{15}}{{.6}^{40}}}{{{9}^{50}}{{.12}^{25}}}=\dfrac{{{2}^{15}}.{{\left( 2.3 \right)}^{40}}}{{{\left( {{3}^{2}} \right)}^{50}}.{{\left( {{2}^{2}}.3 \right)}^{25}}}=\dfrac{{{2}^{15}}{{.2}^{40}}{{.3}^{40}}}{{{3}^{100}}{{.2}^{50}}{{.3}^{25}}}=\dfrac{{{2}^{55}}{{.3}^{40}}}{{{2}^{50}}{{.3}^{125}}}={{2}^{5}}{{.3}^{-85}}$
Suy ra ${{2}^{y}}{{.3}^{x+y}}={{2}^{5}}{{.3}^{-85}}\Leftrightarrow \left\{ \begin{aligned}
& y=5 \\
& x+y=-85 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=5 \\
& x=-90 \\
\end{aligned} \right.$
Vậy $x.y=5.\left( -90 \right)=-450$
$VT={{3}^{x}}{{.6}^{y}}={{3}^{x}}.{{\left( 2.3 \right)}^{y}}={{2}^{y}}.3{{x}^{x+y}}$
$VP=\dfrac{{{2}^{15}}{{.6}^{40}}}{{{9}^{50}}{{.12}^{25}}}=\dfrac{{{2}^{15}}.{{\left( 2.3 \right)}^{40}}}{{{\left( {{3}^{2}} \right)}^{50}}.{{\left( {{2}^{2}}.3 \right)}^{25}}}=\dfrac{{{2}^{15}}{{.2}^{40}}{{.3}^{40}}}{{{3}^{100}}{{.2}^{50}}{{.3}^{25}}}=\dfrac{{{2}^{55}}{{.3}^{40}}}{{{2}^{50}}{{.3}^{125}}}={{2}^{5}}{{.3}^{-85}}$
Suy ra ${{2}^{y}}{{.3}^{x+y}}={{2}^{5}}{{.3}^{-85}}\Leftrightarrow \left\{ \begin{aligned}
& y=5 \\
& x+y=-85 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=5 \\
& x=-90 \\
\end{aligned} \right.$
Vậy $x.y=5.\left( -90 \right)=-450$
Đáp án B.