Đáp án đúng: B
Giải chi tiết:a) \(\left\{ \begin{align} & \left( 2x+4y-1 \right)\sqrt{2x-y-1}=\left( 4x-2y-3 \right)\sqrt{x+2y}\,\,\,\,\,\,\,\,\,\left( 1 \right) \\ & {{x}^{2}}+8x+5-2\left( 3y+2 \right)\sqrt{4x-3y}=2\sqrt{2{{x}^{2}}+5x+2}\,\,\,\,\,\,\left( 2 \right) \\\end{align} \right.\)
\(\left( 1 \right)\Leftrightarrow \left[ 2\left( x+2y \right)-1 \right]\sqrt{2x-y-1}=2\left[ 2\left( 2x-y-1 \right)-1 \right]\sqrt{x+2y}\)
Đặt \(\left\{ \begin{align} & \sqrt{2x-y-1}=a\,\,\left( a\ge 0 \right) \\ & \sqrt{x+2y}=b\,\,\left( b\ge 0 \right) \\\end{align} \right.\) ta có
\(\begin{array}{l}\left( 1 \right) \Leftrightarrow \left( {2{b^2} - 1} \right)a = \left( {2{a^2} - 1} \right)b\\ \Leftrightarrow 2a{b^2} - a - 2{a^2}b + b = 0\\ \Leftrightarrow 2ab\left( {b - a} \right) + \left( {b - a} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}a = b\\2ab = - 1\,\,\left( {vo\,\,nghiem\,\,vi\,\,a;b \ge 0} \right)\end{array} \right.\\ \Leftrightarrow \sqrt {2x - y - 1} = \sqrt {x + 2y} \Leftrightarrow 2x - y - 1 = x + 2y \Leftrightarrow x = 3y + 1\end{array}\)
Thay \(x=3y+1\) vào phương trình (2) ta có \({{x}^{2}}+8x+5-2\left( x+1 \right)\sqrt{3x+1}=2\sqrt{2{{x}^{2}}+5x+2}\)
\(\begin{array}{l} \Leftrightarrow \left[ {{{\left( {x + 1} \right)}^2} - 2\left( {x + 1} \right)\sqrt {3x + 1} + 3x + 1} \right] + \left[ {2x + 1 - 2\sqrt {\left( {2x + 1} \right)\left( {x + 2} \right)} + x + 2} \right] = 0\\ \Leftrightarrow {\left( {x + 1 - \sqrt {3x + 1} } \right)^2} + {\left( {\sqrt {2x + 1} - \sqrt {x + 2} } \right)^2} = 0\\ \Leftrightarrow \left\{ \begin{array}{l}x + 1 = \sqrt {3x + 1} \\\sqrt {2x + 1} = \sqrt {x + 2} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge - 1\\{x^2} + 2x + 1 = 3x + 1\\2x + 1 = x + 2\end{array} \right. \Leftrightarrow x = 1\,\,\left( {tm} \right) \Rightarrow y = 0\end{array}\)
Vậy nghiệm của hệ phương trình là \(\left( x;y \right)=\left( 1;0 \right)\)
b) Ta có:
\(\begin{align} & Q=\frac{11a+11b+12c}{\sqrt{8{{a}^{2}}+56}+\sqrt{8{{b}^{2}}+56}+\sqrt{4{{c}^{2}}+7}} \\ & Q=\frac{11a+11b+12c}{\sqrt{8{{a}^{2}}+8\left( ab+2bc+2ca \right)}+\sqrt{8{{b}^{2}}+8\left( ab+2bc+2ca \right)}+\sqrt{4{{c}^{2}}+ab+2bc+2ca}} \\ & Q=\frac{11a+11b+12c}{\sqrt{\left( 4a+4b \right)\left( 2a+4c \right)}+\sqrt{\left( 4a+4b \right)\left( 2b+4c \right)}+\sqrt{\left( a+2c \right)\left( b+2c \right)}} \\ & Q\overset{Cauchy}{\mathop{\ge }}\,\frac{11a+11b+12c}{\frac{\left( 4a+4b \right)+\left( 2a+4c \right)}{2}+\frac{\left( 4a+4b \right)+\left( 2b+4c \right)}{2}+\frac{\left( a+2c \right)+\left( b+2c \right)}{2}} \\ & Q\ge \frac{2\left( 11a+11b+12c \right)}{11a+11b+12c}=2 \\\end{align}\)
Dấu “=” xảy ra
\(\left\{ \begin{array}{l}4a + 4b = 2a + 4c\\4a + 4b = 2b + 4c\\a + 2c = b + 2c\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = b\\c = \frac{3}{2}b\end{array} \right..\)
Lại có: \(ab+2bc+2ca=7\)
\(\begin{array}{l} \Rightarrow {b^2} + 2b.\frac{3}{2}b + 2b.\frac{3}{2}b = 7\\ \Leftrightarrow {b^2} + 3{b^2} + 3{b^2} = 7\\ \Leftrightarrow {b^2} = 1 \Leftrightarrow b = 1\;\;\left( {do\;\;b \in {Z^ + }} \right).\\ \Rightarrow \left\{ \begin{array}{l}a = b = 1\\c = \frac{3}{2}b = \frac{3}{2}\end{array} \right..\end{array}\)
Vậy \(Min\ \ Q=2\) khi \(a=b=1\) và \(c=\frac{3}{2}.\)
Chọn B
