Câu hỏi của Tuan Minh Do Xuan

Question

Giải các hệ phương trình sau:

a) $\left\{{}\begin{matrix}\left(x^{ }-y\right)^2+y^2=25\\left(x+y\right)^2+x^2=26\end{matrix}\right.$

b) \(\left\{{}\begin{matrix}x-y-xy=2+3\sqrt{2}\x^2+y^2=6\end{...

Nguyễn Thành Trương · Accepted Answer

b)Đặt $S=x+y,P=xy$ thì được:

$\left\{ \begin{align}

& S+P=2+3\sqrt{2} \

& {{S}^{2}}-2P=6 \

\end{align} \right.\Rightarrow {{S}^{2}}+2S+1=11+6\sqrt{2}={{\left( 3+\sqrt{2} \right)}^{2}}$

$\begin{array}{l}
 \Rightarrow \left\{ \begin{array}{l}
S = 2 + \sqrt 2 \
P = 2\sqrt 2 
\end{array} \right. \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2;\sqrt 2 } \right),\left( {\sqrt 2 ;2} \right)} \right\}\
\left\{ \begin{array}{l}
S =  - 4 - \sqrt 2 \
P = 6 + 4\sqrt 2 
\end{array} \right.\left( {VN} \right)
\end{array}
$Câu trả lời: b)Đặt $S=x+y,P=xy$ thì được:

$\left\{ \begin{align}

& S+P=2+3\sqrt{2} \

& {{S}^{2}}-2P=6 \

\end{align} \right.\Rightarrow {{S}^{2}}+2S+1=11+6\sqrt{2}={{\left( 3+\sqrt{2} \right)}^{2}}$

$\begin{array}{l}
 \Rightarrow \left\{ \begin{array}{l}
S = 2 + \sqrt 2 \
P = 2\sqrt 2 
\end{array} \right. \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2;\sqrt 2 } \right),\left( {\sqrt 2 ;2} \right)} \right\}\
\left\{ \begin{array}{l}
S =  - 4 - \sqrt 2 \
P = 6 + 4\sqrt 2 
\end{array} \right.\left( {VN} \right)
\end{array}
$

Nguyễn Thành Trương · Answer

$
c)\left\{ \begin{array}{l}
2{x^2} + xy + 3{y^2} - 2y - 4 = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right.\
 \Leftrightarrow \left\{ \begin{array}{l}
2\left( {2{x^2} + xy + 3{y^2} - 2y - 4} \right) - \left( {3{x^2} + 5{y^2} + 4x - 12} \right) = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right.\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 2xy + {y^2} - 4x - 4y + 4 = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right.\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + y - 2} \right)^2} = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y - 2 = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\
y = 1
\end{array} \right.
$Câu trả lời: $
c)\left\{ \begin{array}{l}
2{x^2} + xy + 3{y^2} - 2y - 4 = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right.\
 \Leftrightarrow \left\{ \begin{array}{l}
2\left( {2{x^2} + xy + 3{y^2} - 2y - 4} \right) - \left( {3{x^2} + 5{y^2} + 4x - 12} \right) = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right.\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 2xy + {y^2} - 4x - 4y + 4 = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right.\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + y - 2} \right)^2} = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y - 2 = 0\
3{x^2} + 5{y^2} + 4x - 12 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\
y = 1
\end{array} \right.
$

Nguyễn Thành Trương · Answer

Xét $x=0$ thì $\left\{ \begin{align}

& 2{{y}^{2}}=25 \

& 2{{y}^{2}}=26 \

\end{align} \right.$(vô nghiệm)

Xét $x\ne 0$, đặt $y=kx$. Chia vế theo vế suy ra:$27{{k}^{2}}-102k-24=0\Leftrightarrow \left[ \begin{align}

& k=4 \

& k=-\frac{2}{9} \

\end{align} \right.$

Từ đó giải ra 4 nghiệm$\left( 1;4 \right),\left( -1;-4 \right),\left( \dfrac{9}{\sqrt{5}};-\dfrac{2}{\sqrt{5}} \right),\left( -\dfrac{9}{\sqrt{5}};-\dfrac{2}{\sqrt{5}} \right)$Câu trả lời: Xét $x=0$ thì $\left\{ \begin{align}