Question

1. \(\left\{{}\begin{matrix}x+xy+y=11\\x^2+y^2-xy-2\left(x+y\right)=-31\end{matrix}\right.\)

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

3. \(\left\{{}\begin{matrix}x^2+4y^2=8\\x+2y=4\end{matrix}\right.\)

4. \(\left\{{}\begin{matrix}2+6y=\frac{x}{y}-\sqrt{x-2y}\\\sqrt{x+\sqrt{x-2y}}=x+3y-2\end{matrix}\right.\)

Answer 1

Câu 1:

HPT \(\Leftrightarrow \left\{\begin{matrix} (x+y)+xy=11\\ (x+y)^2-3xy-2(x+y)=-31\end{matrix}\right.\)

Đặt \(\left\{\begin{matrix} x+y=a\\ xy=b\end{matrix}\right.\) thì hệ trở thành:

\( \left\{\begin{matrix} a+b=11\\ a^2-3b-2a=-31\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} b=11-a\\ a^2-3b-2a+31=0\end{matrix}\right.\)

\(\Rightarrow a^2-3(11-a)-2a+31=0\)

\(\Leftrightarrow a^2+a-2=0\Leftrightarrow (a-1)(a+2)=0\)

\(\Rightarrow \left[\begin{matrix} a=1\\ a=-2\end{matrix}\right.\)

Nếu $a=1\Rightarrow b=11-a=10$

Như vậy $x+y=1; xy=10$

\(\Rightarrow x(1-x)=10\Leftrightarrow x^2-x+10=0\Leftrightarrow (x-\frac{1}{2})^2=-\frac{39}{4}< 0\) (vô lý)

Nếu \(a=-2\Rightarrow b=11-a=13\)

Như vậy $x+y=-2; xy=13$

$\Rightarrow x(-2-x)=13\Leftrightarrow x^2+2x+13=0\Leftrightarrow (x+1)^2=-12< 0$ (vô lý)

Vậy HPT vô nghiệm.

Answer 2

Câu 2:

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

Đặt \(xy=a; x-y=b\) thì hệ trở thành:

\(\left\{\begin{matrix} a-b=-3\\ b^2-b+3a=6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=b-3\\ b^2-b+3a-6=0\end{matrix}\right.\)

\(\Rightarrow b^2-b+3(b-3)-6=0\)

\(\Leftrightarrow b^2+2b-15=0\Leftrightarrow (b-3)(b+5)=0\)

\(\Rightarrow \left[\begin{matrix} b=3\\ b=-5\end{matrix}\right.\)

Nếu $b=3=x-y\Rightarrow a=xy=b-3=0$

\(\Rightarrow (x,y)=(0,-3); (3,0)\)

Nếu \(b=x-y=-5\Rightarrow a=xy=b-3=-8\)

\(\Rightarrow (y-5)y=-8\)

\(\Leftrightarrow y^2-5y+8=0\Leftrightarrow (y-2,5)^2=-1,75< 0\) (vô lý)

Vậy $(x,y)=(0,-3)$ hoặc $(3,0)$

Answer 3

Câu 3:

HPT \(\Leftrightarrow \left\{\begin{matrix} x^2+4y^2=8\\ x=4-2y\end{matrix}\right.\Rightarrow (4-2y)^2+4y^2=8\)

\(\Leftrightarrow 8y^2-16y+8=0\Leftrightarrow y^2-2y+1=0\)

\(\Leftrightarrow (y-1)^2=0\Rightarrow y=1\)

Thay $y=1$ có $x=4-2y=2$

Vậy $(x,y)=(2,1)$

Answer 4

Câu 4:

ĐK: ....................

PT\((1)\Leftrightarrow 6y=\frac{x-2y}{y}-\sqrt{x-2y}\)

Đặt $\sqrt{x-2y}=a(a\geq 0$). Khi đó :

\(6y=\frac{a^2}{y}-a\)

\(\Rightarrow 6y^2=a^2-ay\)

\(\Leftrightarrow 6y^2+ay-a^2=0\)

\(\Leftrightarrow (3y-a)(2y+a)=0\Rightarrow \left[\begin{matrix} 3y=a\\ -2y=a\end{matrix}\right.\)

Nếu \(a=3y=\sqrt{x-2y}\Rightarrow x=9y^2+2y\)

PT\((2)\Rightarrow \sqrt{9y^2+2y+3y}=9y^2+2y+3y-2\)

\(\Leftrightarrow 9y^2+5y-\sqrt{9y^2+5y}-2=0\)

\(\Leftrightarrow (\sqrt{9y^2+5y}-2)(\sqrt{9y^2+5y}+1)=0\)

\(\Rightarrow \sqrt{9y^2+5y}=2\) (TH còn lại dễ thấy vô lý)

\(\Rightarrow 9y^2+5y=4\Leftrightarrow 9y^2+5y-4=0\)

\(\Leftrightarrow (9y-4)(y+1)=0\Rightarrow y=\frac{4}{9}\) hoặc $y=-1$

Thử lại thấy $y=\frac{4}{9}$ thỏa mãn. Kéo theo $x=\frac{8}{3}$ là đáp án duy nhất thỏa mãn.