a) Giải hệ phương trình \(\left\{ \begin{align}  &...

B

- Hướng dẫn giải

Phương pháp giải:

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