1) Giải phương trình\(\left\{ \begin{array}{l}...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

1) Giải phương trình 

\(\left\{ \begin{array}{l}
{x^3} - {y^3} - 6{x^2} + 13x - y = 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
\sqrt {2x + y + 5} - \sqrt {3 - x - y} = \left( {2x - 5} \right)y + 2\,\,\,\,\left( 2 \right)
\end{array} \right.\)

\(\left\{ \begin{align}  & {{x}^{3}}-{{y}^{3}}-6{{x}^{2}}+13x-y=10\text{ }\left( 1 \right) \\  & \sqrt{2x+y+5}-\sqrt{3-x-y}=\left( 2x-5 \right)y+2\text{ }\left( 2 \right) \\ \end{align} \right.\)

Ta có

\(\begin{align}  & \left( 1 \right)\Leftrightarrow {{\left( x-2 \right)}^{3}}-{{y}^{3}}+x-2-y=0 \\  & \Leftrightarrow \left( x-2-y \right)\left[ {{\left( x-2 \right)}^{2}}+y\left( x-2 \right)+{{y}^{2}}+1 \right]=0 \\ \end{align}\)

Vì \({{\left( x-2 \right)}^{2}}+y\left( x-2 \right)+{{y}^{2}}+1={{\left( x-2+\frac{y}{2} \right)}^{2}}+\frac{3{{y}^{2}}}{4}+1>0,\forall x,y\) nên\(\left( 1 \right)\Leftrightarrow x-2-y=0\Leftrightarrow x=y+2\)

Khi đó phương trình (2) trở thành \(\sqrt{3y+9}-\sqrt{1-2y}=\left( 2y-1 \right)y+2\)

Điều kiện: \(-3\le y\le \frac{1}{2}\)

Nếu \(0<y\le \frac{1}{2}\Rightarrow \left\{ \begin{align}  & \sqrt{3y+9}-\sqrt{1-2y}>\sqrt{9}-\sqrt{1}=2 \\ & \left( 2y-1 \right)y\le 0\Rightarrow \left( 2y-1 \right)y+2\le 2 \\\end{align} \right.\Rightarrow VT>VP\) (loại)

Nếu \(-3\le y<0\Rightarrow \left\{ \begin{align}  & \sqrt{3y+9}-\sqrt{1-2y}<2 \\& \left( 2y-1 \right)y+2>2 \\\end{align} \right.\Rightarrow VT<VP\)

Vậy \(y=0\Rightarrow x=2\)

Hệ có nghiệm duy nhất \(\left( x;\ y \right)=\left( 2;\ 0 \right)\)

2) Cho a, b, c là các số thực dương thỏa mãn \({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc=4\) Chứng minh rằng \(2a+b+c\le \frac{9}{2}\)

 Ta có \(4={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+abc={{a}^{2}}+{{\left( b+c \right)}^{2}}+\left( a-2 \right)bc\)

Vì \(a,\ b,\ c>0\) nên \({{a}^{2}}=4-\left( {{b}^{2}}+{{c}^{2}}+abc \right)<4\Rightarrow 0<a<2\Rightarrow \left( a-2 \right)bc\ge \left( a-2 \right).\frac{{{\left( b+c \right)}^{2}}}{4}\)

\(\begin{align}  & \Rightarrow 4\ge {{a}^{2}}+{{\left( b+c \right)}^{2}}+\left( a-2 \right).\frac{{{\left( b+c \right)}^{2}}}{4}={{a}^{2}}+\frac{\left( a+2 \right){{\left( b+c \right)}^{2}}}{4} \\  & \Leftrightarrow 4{{a}^{2}}+\left( a+2 \right){{\left( b+c \right)}^{2}}\le 16 \\  & \Leftrightarrow \left( a+2 \right)\left[ 4\left( a-2 \right)+{{\left( b+c \right)}^{2}} \right]\le 0 \\  & \Rightarrow 4a-8+{{\left( b+c \right)}^{2}}\le 0 \\  & \Rightarrow {{\left( b+c \right)}^{2}}\le 8-4a \\ & \Rightarrow b+c\le 2\sqrt{2-a} \\  & \Rightarrow 2a+b+c=2a+2\sqrt{2-a}=-\frac{1}{2}\left( 4\left( 2-a \right)-4\sqrt{2-a}+1 \right)+\frac{9}{2}=-\frac{1}{2}{{\left( 2\sqrt{2-a}-1 \right)}^{2}}+\frac{9}{2}\le \frac{9}{2} \\ \end{align}\)

Dấu “=” xảy ra 

\(\Leftrightarrow \left\{ \begin{array}{l}
2\sqrt {2 - a} = 1\\
b = c\\
2a + b + c = \frac{9}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = \frac{7}{4}\\
b = c = \frac{1}{2}
\end{array} \right..\)

 Chọn A