Diễn Đàn MathScope - Xem bài viết đơn - Đề thi Olympic 30-4 lớp 11

hanguyen445 · 08-04-2017, 03:23 PM

\[\begin{gathered}
{\text{Giải hệ phương trình }}\left\{ \begin{gathered}
\frac{3}{{\sqrt y }} - \frac{1}{x} = \frac{{5x + \sqrt y }}{{2{x^2} + y}} \hfill \\
\frac{1}{{xy}} + \frac{4}{{\sqrt y }} = \frac{2}{y} + \frac{8}{3} \hfill \\
\end{gathered} \right. \hfill \\
HD\left[ 1 \right] \hfill \\
{\text{Xét PT}}\left( {\text{1}} \right){\text{ dạng đẳng cấp }}x,\sqrt y \to {\text{ du }}x = \sqrt y \to \frac{2}{x} = \frac{{6x}}{{3{x^2}}} \to {\text{ PT }}\left( {\text{1}} \right){\text{ có nhân tử:}}\,{\text{x = }}\sqrt {\text{y}} \hfill \\
{\text{Điều kiện phương trình:}}\left\{ \begin{gathered}
y > 0 \hfill \\
x \ne 0 \hfill \\
\end{gathered} \right. \hfill \\
{\text{PT}}\left( 1 \right) \Leftrightarrow \frac{{3x}}{{\sqrt y }} - 1 = \frac{{5{x^2} + x\sqrt y }}{{2{x^2} + y}} = \frac{{5{{\left( {\frac{x}{{\sqrt y }}} \right)}^2} + \frac{x}{{\sqrt y }}}}{{2.\left( {\frac{x}{{\sqrt y }}} \right) + 1}} \Leftrightarrow 3t - 1 = \frac{{5{t^2} + t}}{{2t + 1}}\left( {t = \frac{x}{{\sqrt y }} \ne - \frac{1}{2}} \right) \hfill \\
\Leftrightarrow \left( {3t - 1} \right)\left( {2t + 1} \right) = 5{t^2} + t \Leftrightarrow {t^2} - 1 = 0 \Leftrightarrow \left[ \begin{gathered}
t = 1 \hfill \\
t = - 1 \hfill \\
\end{gathered} \right. \Rightarrow \left[ \begin{gathered}
x = - \sqrt y \hfill \\
x = \sqrt y \hfill \\
\end{gathered} \right. \hfill \\
\bullet {\text{ Với }}x = \sqrt y > 0 \Rightarrow PT\left( 2 \right) \Leftrightarrow \frac{1}{{{x^3}}} + \frac{4}{x} = \frac{2}{{{x^2}}} + \frac{8}{3} \Leftrightarrow 8{x^3} - 12{x^2} + 6x - 3 = 0 \hfill \\
\Leftrightarrow {\left( {2x - 1} \right)^3} - 2 = 0 \Leftrightarrow x = \frac{{1 + \sqrt[3]{2}}}{2} = \sqrt y \hfill \\
\bullet {\text{ Với }}x = - \sqrt y < 0 \Rightarrow PT\left( 2 \right) \Leftrightarrow \frac{1}{{{x^3}}} - \frac{4}{x} = \frac{2}{{{x^2}}} + \frac{8}{3} \Leftrightarrow 8{x^3} + 12{x^2} + 6x - 3 = 0 \Leftrightarrow {\left( {2x + 1} \right)^3} = 4 \hfill \\
\Leftrightarrow x\frac{{\sqrt[3]{4} - 1}}{2} > 0\left( {{\text{ loai}}} \right) \hfill \\
\bullet {\text{ Kết luân: Vậy nghiệm của hệ }}\left( {x;y} \right) = \left( {\frac{{1 + \sqrt[3]{2}}}{2} = {a_0};{a_0}^2} \right) \hfill \\
\end{gathered} \]

08-04-2017, 03:23 PM	#5
hanguyen445 +Thành Viên+ Tham gia ngày: Oct 2014 Bài gởi: 1 Thanks: 0 Thanked 0 Times in 0 Posts	\[\begin{gathered} {\text{Giải hệ phương trình }}\left\{ \begin{gathered} \frac{3}{{\sqrt y }} - \frac{1}{x} = \frac{{5x + \sqrt y }}{{2{x^2} + y}} \hfill \\ \frac{1}{{xy}} + \frac{4}{{\sqrt y }} = \frac{2}{y} + \frac{8}{3} \hfill \\ \end{gathered} \right. \hfill \\ HD\left[ 1 \right] \hfill \\ {\text{Xét PT}}\left( {\text{1}} \right){\text{ dạng đẳng cấp }}x,\sqrt y \to {\text{ du }}x = \sqrt y \to \frac{2}{x} = \frac{{6x}}{{3{x^2}}} \to {\text{ PT }}\left( {\text{1}} \right){\text{ có nhân tử:}}\,{\text{x = }}\sqrt {\text{y}} \hfill \\ {\text{Điều kiện phương trình:}}\left\{ \begin{gathered} y > 0 \hfill \\ x \ne 0 \hfill \\ \end{gathered} \right. \hfill \\ {\text{PT}}\left( 1 \right) \Leftrightarrow \frac{{3x}}{{\sqrt y }} - 1 = \frac{{5{x^2} + x\sqrt y }}{{2{x^2} + y}} = \frac{{5{{\left( {\frac{x}{{\sqrt y }}} \right)}^2} + \frac{x}{{\sqrt y }}}}{{2.\left( {\frac{x}{{\sqrt y }}} \right) + 1}} \Leftrightarrow 3t - 1 = \frac{{5{t^2} + t}}{{2t + 1}}\left( {t = \frac{x}{{\sqrt y }} \ne - \frac{1}{2}} \right) \hfill \\ \Leftrightarrow \left( {3t - 1} \right)\left( {2t + 1} \right) = 5{t^2} + t \Leftrightarrow {t^2} - 1 = 0 \Leftrightarrow \left[ \begin{gathered} t = 1 \hfill \\ t = - 1 \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered} x = - \sqrt y \hfill \\ x = \sqrt y \hfill \\ \end{gathered} \right. \hfill \\ \bullet {\text{ Với }}x = \sqrt y > 0 \Rightarrow PT\left( 2 \right) \Leftrightarrow \frac{1}{{{x^3}}} + \frac{4}{x} = \frac{2}{{{x^2}}} + \frac{8}{3} \Leftrightarrow 8{x^3} - 12{x^2} + 6x - 3 = 0 \hfill \\ \Leftrightarrow {\left( {2x - 1} \right)^3} - 2 = 0 \Leftrightarrow x = \frac{{1 + \sqrt[3]{2}}}{2} = \sqrt y \hfill \\ \bullet {\text{ Với }}x = - \sqrt y < 0 \Rightarrow PT\left( 2 \right) \Leftrightarrow \frac{1}{{{x^3}}} - \frac{4}{x} = \frac{2}{{{x^2}}} + \frac{8}{3} \Leftrightarrow 8{x^3} + 12{x^2} + 6x - 3 = 0 \Leftrightarrow {\left( {2x + 1} \right)^3} = 4 \hfill \\ \Leftrightarrow x\frac{{\sqrt[3]{4} - 1}}{2} > 0\left( {{\text{ loai}}} \right) \hfill \\ \bullet {\text{ Kết luân: Vậy nghiệm của hệ }}\left( {x;y} \right) = \left( {\frac{{1 + \sqrt[3]{2}}}{2} = {a_0};{a_0}^2} \right) \hfill \\ \end{gathered} \]