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Câu hỏi: Cho hai số thực $x, y$ thỏa mãn điều kiện $3{{\left( x+y \right)}^{2}}+5\left( x-y \right){{~}^{2}}=4$. Hỏi có bao nhiêu giá trị nguyên của mthỏa mãn $m\left( 2xy+1 \right)=1010{{(x{{~}^{2}}+{{y}^{2}}~)}^{2}}+1010{{({{x}^{2}}~-{{y}^{2}})}^{2}}?~$
A. 235
B. 1175
C. 1176
D. 236
A. 235
B. 1175
C. 1176
D. 236
Cách giải:
Theo bài ra ta có:
$3{{\left( x+y \right)}^{2}}+5{{\left( x-y \right)}^{2}}=4$
$\Leftrightarrow 3{{x}^{2}}+3{{y}^{2}}+6xy+5{{x}^{2}}+5{{y}^{2}}-10xy=4$
$\Leftrightarrow 8\left( {{x}^{2}}+{{y}^{2}} \right)=4xy+4$
$\begin{aligned}
& \Leftrightarrow 2\left( {{x}^{2}}+{{y}^{2}} \right)=xy+1 \\
& \Leftrightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{xy+1}{2} \left( 1 \right) \\
\end{aligned}$
Mặt khác ta lại có ${{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}={{\left( x+{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}\left( 2 \right)$
Đặt$\left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}=u \\
& xy= \\
\end{aligned} \right., $ từ (1)và (2) ta có$ \left\{ \begin{aligned}
& u=\dfrac{1}{2}\left( v+1 \right) \\
& {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{u}^{2}}-4{{v}^{2}}=\dfrac{1}{4}{{\left( v+1 \right)}^{2}}-4{{v}^{2}} \\
\end{aligned} \right.$
$\Rightarrow {{u}^{2}}=\dfrac{1}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}-4{{v}^{2}}=-\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}$
Do$\left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}\ge 0 \\
& {{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}\ge 0\Leftrightarrow \\
& {{x}^{2}}+{{y}^{2}}\ge 2xy \\
\end{aligned} \right.\left\{ \begin{aligned}
& \dfrac{1}{2}\left( v+1 \right)\ge 0 \\
& -\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}\ge 0\Leftrightarrow \\
& \dfrac{1}{2}\left( v+1 \right)\ge 2v \\
\end{aligned} \right.\left\{ \begin{aligned}
& v\ge -1 \\
& -\dfrac{1}{5}\le v\le \dfrac{1}{3}\Leftrightarrow -\dfrac{1}{5}\le v\le \dfrac{1}{3} \\
& v\le \dfrac{1}{3} \\
\end{aligned} \right.$
Thay ${{x}^{2}}+{{y}^{2}}=u,xy=v$ và $u=\dfrac{1}{2}\left( v+1 \right),{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=-\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+1$ ta có:
$m\left( 2v+1 \right)=1010{{\left[ \dfrac{1}{2}\left( v+ \right) \right]}^{2}}+1010\left[ -\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4} \right]$
$\Leftrightarrow m\left( 2v+1 \right)=1010\left[ \dfrac{1}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}-\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4} \right]$
$\Leftrightarrow m\left( 2v+1 \right)=1010\left[ -\dfrac{7}{2}{{v}^{2}}+v+\dfrac{1}{2} \right]$
$\Leftrightarrow m=\dfrac{505\left( -7{{v}^{2}}+2v+1 \right)}{2v+1}$
Xét hàm số $f\left( v \right)=\dfrac{-7{{v}^{2}}+2v+1}{2v+1}$ trên $\left[ -\dfrac{1}{5};\dfrac{1}{3} \right]$ ta có:
$f'\left( v \right)=\dfrac{\left( -14v+2 \right).\left( 2v+1 \right)-\left( -7{{v}^{2}}+2v+1 \right).2}{{{\left( 2v+1 \right)}^{2}}}$
$f'\left( v \right)=\dfrac{-28{{v}^{2}}-14v+4v+2+14{{v}^{2}}-4v-2}{{{\left( 2v+1 \right)}^{2}}}$
$f'\left( v \right)=\dfrac{-14{{v}^{2}}-14v}{{{\left( 2v+1 \right)}^{2}}}$
$f'\left( v \right)=0\Leftrightarrow \left[ \begin{aligned}
& v=0\in \left[ -\dfrac{1}{5};\dfrac{1}{3} \right] \\
& v=-1\notin \left[ -\dfrac{1}{5};\dfrac{1}{3} \right] \\
\end{aligned} \right.$
Ta có : $f\left( 0 \right)=1,f\left( -\dfrac{1}{5} \right)=\dfrac{8}{15},f\left( \dfrac{1}{3} \right)=\dfrac{8}{15}$
$\Rightarrow \dfrac{8}{15}\le f\left( v \right)\le 1\forall v\in \left[ -\dfrac{1}{5};\dfrac{1}{3} \right]\Rightarrow \dfrac{808}{3}\le m\le 505$
Mà $m\in \mathbb{Z}\Rightarrow m\in \left\{ 270;271;...;505 \right\}.$
Vậy số giá trị của mthỏa mãn yêu cầu bài toán là $505-270+1=236.~$
Theo bài ra ta có:
$3{{\left( x+y \right)}^{2}}+5{{\left( x-y \right)}^{2}}=4$
$\Leftrightarrow 3{{x}^{2}}+3{{y}^{2}}+6xy+5{{x}^{2}}+5{{y}^{2}}-10xy=4$
$\Leftrightarrow 8\left( {{x}^{2}}+{{y}^{2}} \right)=4xy+4$
$\begin{aligned}
& \Leftrightarrow 2\left( {{x}^{2}}+{{y}^{2}} \right)=xy+1 \\
& \Leftrightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{xy+1}{2} \left( 1 \right) \\
\end{aligned}$
Mặt khác ta lại có ${{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}={{\left( x+{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}\left( 2 \right)$
Đặt$\left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}=u \\
& xy= \\
\end{aligned} \right., $ từ (1)và (2) ta có$ \left\{ \begin{aligned}
& u=\dfrac{1}{2}\left( v+1 \right) \\
& {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{u}^{2}}-4{{v}^{2}}=\dfrac{1}{4}{{\left( v+1 \right)}^{2}}-4{{v}^{2}} \\
\end{aligned} \right.$
$\Rightarrow {{u}^{2}}=\dfrac{1}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}-4{{v}^{2}}=-\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}$
Do$\left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}\ge 0 \\
& {{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}\ge 0\Leftrightarrow \\
& {{x}^{2}}+{{y}^{2}}\ge 2xy \\
\end{aligned} \right.\left\{ \begin{aligned}
& \dfrac{1}{2}\left( v+1 \right)\ge 0 \\
& -\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}\ge 0\Leftrightarrow \\
& \dfrac{1}{2}\left( v+1 \right)\ge 2v \\
\end{aligned} \right.\left\{ \begin{aligned}
& v\ge -1 \\
& -\dfrac{1}{5}\le v\le \dfrac{1}{3}\Leftrightarrow -\dfrac{1}{5}\le v\le \dfrac{1}{3} \\
& v\le \dfrac{1}{3} \\
\end{aligned} \right.$
Thay ${{x}^{2}}+{{y}^{2}}=u,xy=v$ và $u=\dfrac{1}{2}\left( v+1 \right),{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=-\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+1$ ta có:
$m\left( 2v+1 \right)=1010{{\left[ \dfrac{1}{2}\left( v+ \right) \right]}^{2}}+1010\left[ -\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4} \right]$
$\Leftrightarrow m\left( 2v+1 \right)=1010\left[ \dfrac{1}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4}-\dfrac{15}{4}{{v}^{2}}+\dfrac{1}{2}v+\dfrac{1}{4} \right]$
$\Leftrightarrow m\left( 2v+1 \right)=1010\left[ -\dfrac{7}{2}{{v}^{2}}+v+\dfrac{1}{2} \right]$
$\Leftrightarrow m=\dfrac{505\left( -7{{v}^{2}}+2v+1 \right)}{2v+1}$
Xét hàm số $f\left( v \right)=\dfrac{-7{{v}^{2}}+2v+1}{2v+1}$ trên $\left[ -\dfrac{1}{5};\dfrac{1}{3} \right]$ ta có:
$f'\left( v \right)=\dfrac{\left( -14v+2 \right).\left( 2v+1 \right)-\left( -7{{v}^{2}}+2v+1 \right).2}{{{\left( 2v+1 \right)}^{2}}}$
$f'\left( v \right)=\dfrac{-28{{v}^{2}}-14v+4v+2+14{{v}^{2}}-4v-2}{{{\left( 2v+1 \right)}^{2}}}$
$f'\left( v \right)=\dfrac{-14{{v}^{2}}-14v}{{{\left( 2v+1 \right)}^{2}}}$
$f'\left( v \right)=0\Leftrightarrow \left[ \begin{aligned}
& v=0\in \left[ -\dfrac{1}{5};\dfrac{1}{3} \right] \\
& v=-1\notin \left[ -\dfrac{1}{5};\dfrac{1}{3} \right] \\
\end{aligned} \right.$
Ta có : $f\left( 0 \right)=1,f\left( -\dfrac{1}{5} \right)=\dfrac{8}{15},f\left( \dfrac{1}{3} \right)=\dfrac{8}{15}$
$\Rightarrow \dfrac{8}{15}\le f\left( v \right)\le 1\forall v\in \left[ -\dfrac{1}{5};\dfrac{1}{3} \right]\Rightarrow \dfrac{808}{3}\le m\le 505$
Mà $m\in \mathbb{Z}\Rightarrow m\in \left\{ 270;271;...;505 \right\}.$
Vậy số giá trị của mthỏa mãn yêu cầu bài toán là $505-270+1=236.~$
Đáp án D.