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Câu hỏi: Biết phưong trình ${{\log }_{2018}}\left( \dfrac{2}{\sqrt{x}}+\dfrac{1}{x} \right)=2{{\log }_{2019}}\left( \dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}} \right)$ có nghiệm duy nhất $x=a+b\sqrt{c}$ trong đó a; b là những số nguyên; c là số nguyên tố. Khi đó a + b + c bằng
A. 5.
B. 1.
C. 7.
D. 3.
A. 5.
B. 1.
C. 7.
D. 3.
Điều kiện x > 1.
${{\log }_{2018}}\left( \dfrac{2}{\sqrt{x}}+\dfrac{1}{x} \right)=2{{\log }_{2019}}\left( \dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}} \right)\Leftrightarrow {{\log }_{2018}}\left( \dfrac{2\sqrt{x}+1}{x} \right)={{\log }_{\sqrt{2019}}}\left( \dfrac{x-1}{2\sqrt{x}} \right)=t$
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{2\sqrt{x}+1}{x}={{2018}^{t}} \\
& \dfrac{x-1}{2\sqrt{x}}={{\left( \sqrt{2019} \right)}^{t}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2\sqrt{x}+1=x{{.2018}^{t}} \left( 1 \right) \\
& x-1=2\sqrt{x}.{{\left( \sqrt{2019} \right)}^{t}} \left( 2 \right) \\
\end{aligned} \right.$
Lấy (1) cộng (2) vế theo vế ta được $x+2\sqrt{x}=x{{.2018}^{t}}+2\sqrt{x}.{{\left( \sqrt{2019} \right)}^{t}}$
$\Leftrightarrow x\left( {{2018}^{t}}-1 \right)+2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]=0\left( 3 \right)$
+ TH1:
Xét $t>0\Rightarrow \left\{ \begin{aligned}
& {{2018}^{t}}>{{2018}^{0}}=1 \\
& {{\left( \sqrt{2019} \right)}^{t}}>{{\left( \sqrt{2019} \right)}^{0}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{2018}^{t}}-1>0 \\
& {{\left( \sqrt{2019} \right)}^{t}}-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x.\left( {{2018}^{t}}-1 \right)>0 \\
& 2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]>0 \\
\end{aligned} \right.$
$\Rightarrow x\left( {{2018}^{t}}-1 \right)+2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]>0;\forall t>0\Rightarrow $ (3) vô nghiệm.
+ TH2:
Xét $t<0\Rightarrow \left\{ \begin{aligned}
& {{2018}^{t}}<{{2018}^{0}}=1 \\
& {{\left( \sqrt{2019} \right)}^{t}}<{{\left( \sqrt{2019} \right)}^{0}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{2018}^{t}}-1<0 \\
& {{\left( \sqrt{2019} \right)}^{t}}-1<0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x.\left( {{2018}^{t}}-1 \right)<0 \\
& 2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]<0 \\
\end{aligned} \right.$
$\Rightarrow x\left( {{2018}^{t}}-1 \right)+2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]<0;\forall t<0\Rightarrow $ (3) vô nghiệm.
+ TH3: Xét thấy $t=0$ thỏa mãn phương trình (3).
Từ các trường hợp trên, ta có (3) $\Leftrightarrow t=0$.
Thế vào (1), ta có $2\sqrt{x}+1=x\Leftrightarrow x-2\sqrt{x}-1=0\Leftrightarrow \sqrt{x}=1+\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}$
Vậy $a+b+c=7.$
${{\log }_{2018}}\left( \dfrac{2}{\sqrt{x}}+\dfrac{1}{x} \right)=2{{\log }_{2019}}\left( \dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}} \right)\Leftrightarrow {{\log }_{2018}}\left( \dfrac{2\sqrt{x}+1}{x} \right)={{\log }_{\sqrt{2019}}}\left( \dfrac{x-1}{2\sqrt{x}} \right)=t$
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{2\sqrt{x}+1}{x}={{2018}^{t}} \\
& \dfrac{x-1}{2\sqrt{x}}={{\left( \sqrt{2019} \right)}^{t}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2\sqrt{x}+1=x{{.2018}^{t}} \left( 1 \right) \\
& x-1=2\sqrt{x}.{{\left( \sqrt{2019} \right)}^{t}} \left( 2 \right) \\
\end{aligned} \right.$
Lấy (1) cộng (2) vế theo vế ta được $x+2\sqrt{x}=x{{.2018}^{t}}+2\sqrt{x}.{{\left( \sqrt{2019} \right)}^{t}}$
$\Leftrightarrow x\left( {{2018}^{t}}-1 \right)+2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]=0\left( 3 \right)$
+ TH1:
Xét $t>0\Rightarrow \left\{ \begin{aligned}
& {{2018}^{t}}>{{2018}^{0}}=1 \\
& {{\left( \sqrt{2019} \right)}^{t}}>{{\left( \sqrt{2019} \right)}^{0}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{2018}^{t}}-1>0 \\
& {{\left( \sqrt{2019} \right)}^{t}}-1>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x.\left( {{2018}^{t}}-1 \right)>0 \\
& 2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]>0 \\
\end{aligned} \right.$
$\Rightarrow x\left( {{2018}^{t}}-1 \right)+2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]>0;\forall t>0\Rightarrow $ (3) vô nghiệm.
+ TH2:
Xét $t<0\Rightarrow \left\{ \begin{aligned}
& {{2018}^{t}}<{{2018}^{0}}=1 \\
& {{\left( \sqrt{2019} \right)}^{t}}<{{\left( \sqrt{2019} \right)}^{0}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{2018}^{t}}-1<0 \\
& {{\left( \sqrt{2019} \right)}^{t}}-1<0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x.\left( {{2018}^{t}}-1 \right)<0 \\
& 2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]<0 \\
\end{aligned} \right.$
$\Rightarrow x\left( {{2018}^{t}}-1 \right)+2\sqrt{x}\left[ {{\left( \sqrt{2019} \right)}^{t}}-1 \right]<0;\forall t<0\Rightarrow $ (3) vô nghiệm.
+ TH3: Xét thấy $t=0$ thỏa mãn phương trình (3).
Từ các trường hợp trên, ta có (3) $\Leftrightarrow t=0$.
Thế vào (1), ta có $2\sqrt{x}+1=x\Leftrightarrow x-2\sqrt{x}-1=0\Leftrightarrow \sqrt{x}=1+\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}$
Vậy $a+b+c=7.$
Đáp án C.