Chủ đề này có 9 trả lời

[Durkein]

Giải các hệ phương trình sau:

1) $\left\{\begin{matrix} x^{2} + xy + 2y = 2y^{2} + 2x \\ y\sqrt{x-y+1} + x = 2  \end{matrix}\right.$

 

2) $\left\{\begin{matrix} x^{2} + y^{2} + \frac{2xy}{x+y} = 1 \\ \sqrt{x+y} = x^{2} - y \end{matrix}\right.$

 

3) $\left\{\begin{matrix} 1 + x^{3}y^{3} = 19x^{3} \\ y + xy^{2} = -6x^{2} \end{matrix}\right.$

 

4) $\left\{\begin{matrix} \sqrt{3x}(1+\frac{1}{x+y}) = 2 \\ \sqrt{7y}(1-\frac{1}{x+y}) = 4\sqrt{2} \end{matrix}\right.$

 

5) $\left\{\begin{matrix} x^{2} + y^{2} + xy + 1 = 4y \\ y(x+y)^{2} = 2x^{2} + 7y + 2 \end{matrix}\right.$

 

6) $\left\{\begin{matrix} y+xy^{2}=6x^{2} \\ 1+x^{2}y^{2}=5x^{2} \end{matrix}\right.$

 

7) $\left\{\begin{matrix} (x+y)(1+\frac{1}{xy})=5 \\ (x^{2}+y^{2})(1+\frac{1}{x^{2}y^{2}})=49 \end{matrix}\right.$

 

8) $\left\{\begin{matrix} x^{2}+y+x^{3}y+xy^{2}+xy=-\frac{5}{4} \\ x^{4}+y^{2}+xy(1+2x)=-\frac{5}{4} \end{matrix}\right.$

 

9) $\left\{\begin{matrix} \sqrt{7x+y} + \sqrt{2x+y} = 5 \\ \sqrt{2x+y}+x-y=2 \end{matrix}\right.$

 

10) $\left\{\begin{matrix} (4x^{2}+1)x+(y-3)\sqrt{5-2y}=0 \\ 4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right.$

[BaNam]

Giải các hệ phương trình sau:

1) $\left\{\begin{matrix} x^{2} + xy + 2y = 2y^{2} + 2x \\ y\sqrt{x-y+1} + x = 2  \end{matrix}\right.$

$x^{2}+xy+2y=2y^{2}+2x$
$\Leftrightarrow$ $(x^{2}+xy-y^{2})-2(x-y)=0$
$\Leftrightarrow$ $(x-y)(x+2y)-2(x-y)=0$
$\Leftrightarrow$ $(x-y)(x+2y-2)=0$
$+)$TH1: $x=y$ Thế vào $y\sqrt{x-y+1}+x=2$ ta được 

     $x+x=2$ $\Rightarrow$ $x=y=1$
$+)$ TH2: $x=2-2y$ Thế vào $y\sqrt{x-y+1}+x=2$  ta được:

  $y\sqrt{3-3y}+2-2y=2$
  $\Leftrightarrow$ $y\sqrt{3-3y}-2y=0$
  $\Rightarrow$ $y=0$ $\Rightarrow$ $x=2$

hoặc $\sqrt{3-3y}=2$ $\Rightarrow$ $y=-\frac{1}{3}$; $x=\frac{8}{3}$
Hệ phương trình có 3 nghiệm $(x;y)$ là $(1;1)$ $(2;0)$ $(\frac{8}{3};-\frac{1}{3})$
 

[BaNam]

2) $\left\{\begin{matrix} x^{2} + y^{2} + \frac{2xy}{x+y} = 1 \\ \sqrt{x+y} = x^{2} - y \end{matrix}\right.$

Đk: $x+y>0$
Xét: $x^{2}+y^{2}+\frac{2xy}{x+y}=1$

$\Leftrightarrow$ $(x+y)^{2}-2xy+\frac{2xy}{x+y}-1=0$
$\Leftrightarrow$ $(x+y-1)(x+y+1)-2xy(1-\frac{1}{x+y})=0$
$\Leftrightarrow$ $(x+y-1)(x+y+1)-2xy\frac{x+y-1}{x+y}=0$
$\Rightarrow$$\left[ \begin{array}{ll} x+y-1=0 \color{red}{(1)}\\x+y+1-\frac{2xy}{x+y}=0  \color{red}{(2)}\end{array} \right.$

 

$1$ $\Rightarrow$ Thế vào: $\sqrt{x+y}=x^{2}-y$ ta được:

   $1=x^{2}+x-1$ $\Leftrightarrow$ $x^{2}+x-2=0$ $\Rightarrow$ $\left[ \begin{array}{II} x=1 \Rightarrow y=0 \\x=-2 \Rightarrow y=3 \end{array} \right.$
 

 

$2$ $\Rightarrow$ $(x+y)^{2}+x+y-2xy=0$ (vì $x+y>0$
$\Rightarrow$ $x^{2}+y^{2}+x+y=0$ ( vô lý vì $x+y>0$ và $x^{2}+y^{2}\geq0$)

Hệ phương trình có 2 nghiệm $(x;y)$ là: $(1;0)$ $(-2;3)$

[BaNam]

3) $\left\{\begin{matrix} 1 + x^{3}y^{3} = 19x^{3} \\ y + xy^{2} = -6x^{2} \end{matrix}\right.$

$\Leftrightarrow$ $ \left\{\begin{matrix} (1+xy)[(xy)^{2}-xy+1]=\frac{19x}{6}.(6x^{2}) \\ y(1+xy)=-6x^{2} \end{matrix}\right. $

$\Rightarrow$ $(1+xy)[(xy)^{2}-xy+1]=-\frac{19x}{6}.y(1+xy)$
$\Rightarrow$ $(1+xy)[(xy)^{2}-xy+1+\frac{19xy}{6}]=0$
$\Rightarrow$ $\left[ \begin{array}{ll} (xy)^{2}-xy+1+\frac{19xy}{6}=0 \\ 1+xy=0\end{array} \right.$

$TH1$: $6[(xy)^{2}-xy+1]+19xy=0$
$\Rightarrow$ $6(xy)^{2}+13xy+6=0$ 
$\Rightarrow$ $\left[ \begin{array}{ll}
xy=-\frac{2}{3} \Rightarrow x=\frac{1}{3};y=-2\\
xy=-\frac{3}{2} \Rightarrow x=-\frac{1}{2};y=3
\end{array} \right.$
$TH2$: $xy+1=0$ $\Rightarrow$ $19x^{3}=0$ $\Rightarrow$ $x=0$ (Vô lý)

Vậy hệ có 2 cặp nghiệm $(x;y)$ là $(\frac{1}{3};-2)$ $(-\frac{1}{2};3)$

[BaNam]

8) $\left\{\begin{matrix} x^{2}+y+x^{3}y+xy^{2}+xy=-\frac{5}{4} \\ x^{4}+y^{2}+xy(1+2x)=-\frac{5}{4} \end{matrix}\right.$

$\Rightarrow$ $\left\{\begin{matrix} x^{2}+y+xy(x^{2}+y)+xy=-\frac{5}{4} \\ (x^{2}+y)^{2}+xy=-\frac{5}{4} \end{matrix}\right.$

$\Leftrightarrow$ $\left\{\begin{matrix} (x^{2}+y)(xy+1)+xy=-\frac{5}{4} \\ (x^{2}+y)^{2}+xy=-\frac{5}{4} \end{matrix}\right.$
$\Rightarrow$ $(x^{2}+y)^{2}-(x^{2}+y)(xy+1)=0$
$\Leftrightarrow$ $(x^{2}+y)(x^{2}+y-xy-1)=0$
$\Leftrightarrow$ $\left[ \begin{array}{ll} x^{2}+y=0 \\ x^{2}+y-xy-1=0 \end{array} \right.$
$+)$ TH1: $x^{2}+y=0$ $\Rightarrow$ $xy=-\frac{5}{4}$
$\Rightarrow$ $x^{3}=\frac{5}{4}$ 
$\Rightarrow$ $x=\sqrt[3]{\frac{5}{4}}$ ; $y=-x^{2}=(\sqrt[3]{\frac{5}{4}})^{2}$
$+)$ TH2: $x^{2}+y=xy+1$ $\Rightarrow$ $xy+1+xy(xy+1)+xy=-\frac{5}{4}$
$\Leftrightarrow$ $(xy)^{2}+3xy+\frac{9}{4}=0$
$\Leftrightarrow$ $4(xy)^{2}+12xy+9=0$ 
$\Rightarrow$ $\left[\begin{array}{ll} xy=4 & (1)\\ xy=-\frac{3}{2} & (2)\end{array} \right.$ 
  $-(1)$ $(x^{2}+y)^{2}=-xy-\frac{5}{4} \geq 0$ $\Rightarrow$ $xy \leq -\frac{5}{4}$ $\Rightarrow$ (loại)

  $-(2)$ $\Rightarrow$ $\left\{\begin{matrix} y=-\frac{3}{2x} \\ x^{2}+y=xy+1\end{matrix} \right. $
            $\Rightarrow$ $x^{2}-\frac{3}{2x}=-\frac{1}{2}$
            $\Rightarrow$ $2x^{3}-3x+1=0$
            $\Rightarrow$ $\left[\begin{array}{ll} x=1 \\ x= \frac{-1\pm\sqrt{3}}{2} \end{array} \right. \rightarrow \left[\begin{array}{ll} y=-\frac{3}{2} \\ y=-\frac{3}{-1 \pm \sqrt{3}} \end{array} \right.$

Vậy hệ có 4 cặp nghiệm $(x;y)$ là: $(\sqrt[3]{\frac{5}{4}};\sqrt[3]{(\frac{5}{4})^{2}}$ $(1;-\frac{3}{2})$ $(\frac{-1\pm\sqrt{3}}{2};-\frac{3}{-1\pm\sqrt{3}})$

[BaNam]

5) $\left\{\begin{matrix} x^{2} + y^{2} + xy + 1 = 4y \\ y(x+y)^{2} = 2x^{2} + 7y + 2 \end{matrix}\right.$

Xét $y=0$ không phải là nghiệm của hệ 
Xét $y\neq0$ $\Rightarrow$ $\left\{\begin{matrix} \frac{x^{2}}{y}+\frac{1}{y}+x+y=4 \\ (x+y)^{2}=2\frac{x^{2}}{y}+2\frac{1}{y}+7 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} \frac{x^{2}}{y}+x+y=4 \\ (x+y)^{2}-2(\frac{x^{2}}{y}+\frac{1}{y})=7 \end{matrix}\right.$
Đặt $a=x+y$ ; $b=\frac{x^{2}}{y}+\frac{1}{y}$
$\Rightarrow$ $\left\{\begin{matrix} a+b=4 \\ a^{2}-2b=7 \end{matrix}\right.$
$\Rightarrow$ $a^{2}+2a-15=0$
$\Leftrightarrow$ $\left[\begin{array}{ll} a=-5 \\ a=3 \end{array}\right. \rightarrow \left[\begin{array}{ll} b=9 \\ b=1 \end{array}\right.$
$+)$ $TH1$: $\left\{\begin{matrix} x+y=-5 \\ \frac{x^{2}}{y}+\frac{1}{y}=9 \end{matrix}\right.$
          $\Rightarrow$ $\left\{\begin{matrix} x+y=-5 \\ x^{2}+1=9y \end{matrix}\right.$
          $\Rightarrow$ $\left\{\begin{matrix} y=-x-5 \\ x^{2}+9x+46=0 \end{matrix}\right.$
          $\Rightarrow$ Hệ vô nghiệm.
$+)$ $TH2$: $\left\{\begin{matrix} x+y=3 \\ \frac{x^{2}}{y}+\frac{1}{y}=1 \end{matrix}\right.$
          $\Rightarrow$ $\left\{\begin{matrix} x+y=3 \\ x^{2}+1=y \end{matrix}\right.$
          $\Rightarrow$ $\left\{\begin{matrix} y=3-x \\ x^{2}+1=3-x \end{matrix}\right.$
          $\Rightarrow$ $\left\{\begin{matrix} y=3-x \\ x^{2}+x-2=0 \end{matrix}\right.$

          $\Rightarrow$ $\left[\begin{array}{ll} x=1 \\ x=-2 \end{array}\right. \Rightarrow \left[\begin{array}{ll} y=2 \\ y=5 \end{array}\right. $
          $\Rightarrow$ Hệ có 2 cặp nghiệm $(x;y)$ là: $(1;2)$ $(-2;5)$

[BaNam]

6) $\left\{\begin{matrix} y+xy^{2}=6x^{2} \\ 1+x^{2}y^{2}=5x^{2} \end{matrix}\right.$

Xét $x=0$ không phải là nghiệm của hệ 
Với $x\neq0$ $\Rightarrow$ $\left\{\begin{matrix} y\frac{1}{x^{2}}+y^{2}.\frac{1}{x}=6 \\ \frac{1}{x^{2}}+y^{2}=5 \end{matrix}\right.$
$\Rightarrow$ Đặt $a=\frac{1}{x}$ $y=b$
$\Rightarrow$ $\left\{\begin{matrix} a^{2}b+ab^{2}=6 \\ a^{2}+b^{2}=5 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} (ab)(a+b)=6 \\ (a+b)^{2}-2ab=5 \end{matrix}\right.$
Đặt $u=a+b$ $v=ab$ 
$\rightarrow$ $\left\{\begin{matrix} uv=6 \\ u^{2}-2v=5 \end{matrix}\right.$
$\rightarrow$ $u^{2}-2.\frac{6}{u}=5$ 
$\leftrightarrow$ $u^{3}-5u-12=0$
$\leftrightarrow$ $(u-3)(u^{2}+3u+4)=0$ 
$\leftrightarrow$ $u=3$ (vì $u^{2}+3u+4>0 \forall x) $ $\rightarrow$ $v=2$
$\rightarrow$ $\left\{\begin{matrix} a+b=3 \\ ab=2 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} a=3-b \\ (3-b)b=2 \end{matrix}\right.$
$\rightarrow$ $b^{2}-3b+2=0$
$\rightarrow$ $\left\{\begin{array}{ll} b=1 \\ b=2 \end{array}\right. \rightarrow \left\{\begin{array}{ll} a=2 \\ a=1 \end{array}\right.$
$+)$ TH1: $\left\{\begin{matrix} a=2 \\ b=1 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} \frac{1}{x}=2 \\ y=1 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} x=\frac{1}{2} \\ y=1 \end{matrix}\right.$
$+)$ TH2: $\left\{\begin{matrix} a=1 \\ b=2 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} \frac{1}{x}=1 \\ y=2 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} x=1 \\ y=2 \end{matrix}\right.$
Vậy hệ có 2 cặp nghiệm $(x;y)$ là $(\frac{1}{2};1)$ ; $(1;2)$

[BaNam]

6) $\left\{\begin{matrix} y+xy^{2}=6x^{2} \\ 1+x^{2}y^{2}=5x^{2} \end{matrix}\right.$

Xét $x=0$ không phải là nghiệm của hệ 
Với $x\neq0$ $\Rightarrow$ $\left\{\begin{matrix} y\frac{1}{x^{2}}+y^{2}.\frac{1}{x}=6 \\ \frac{1}{x^{2}}+y^{2}=5 \end{matrix}\right.$
$\Rightarrow$ Đặt $a=\frac{1}{x}$ $y=b$
$\Rightarrow$ $\left\{\begin{matrix} a^{2}b+ab^{2}=6 \\ a^{2}+b^{2}=5 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} (ab)(a+b)=6 \\ (a+b)^{2}-2ab=5 \end{matrix}\right.$
Đặt $u=a+b$ $v=ab$ 
$\rightarrow$ $\left\{\begin{matrix} uv=6 \\ u^{2}-2v=5 \end{matrix}\right.$
$\rightarrow$ $u^{2}-2.\frac{6}{u}=5$ 
$\leftrightarrow$ $u^{3}-5u-12=0$
$\leftrightarrow$ $(u-3)(u^{2}+3u+4)=0$ 
$\leftrightarrow$ $u=3$ (vì $u^{2}+3u+4>0 \forall x) $ $\rightarrow$ $v=2$
$\rightarrow$ $\left\{\begin{matrix} a+b=3 \\ ab=2 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} a=3-b \\ (3-b)b=2 \end{matrix}\right.$
$\rightarrow$ $b^{2}-3b+2=0$
$\rightarrow$ $\left\{\begin{array}{ll} b=1 \\ b=2 \end{array}\right. \rightarrow \left\{\begin{array}{ll} a=2 \\ a=1 \end{array}\right.$
$+)$ TH1: $\left\{\begin{matrix} a=2 \\ b=1 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} \frac{1}{x}=2 \\ y=1 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} x=\frac{1}{2} \\ y=1 \end{matrix}\right.$
$+)$ TH2: $\left\{\begin{matrix} a=1 \\ b=2 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} \frac{1}{x}=1 \\ y=2 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} x=1 \\ y=2 \end{matrix}\right.$
Vậy hệ có 2 cặp nghiệm $(x;y)$ là $(\frac{1}{2};1)$ ; $(1;2)$

[BaNam]

9) $\left\{\begin{matrix} \sqrt{7x+y} + \sqrt{2x+y} = 5 \\ \sqrt{2x+y}+x-y=2 \end{matrix}\right.$

Đk: $7x+y\geq0$ ; $2x+y\geq 0$
$\left\{\begin{matrix} \sqrt{7x+y}=5-\sqrt{2x+y} \\ \sqrt{2x+y}+x-y=2 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} 5-\sqrt{2x+y} \geq0 \\ 7x+y=25+2x+y-10\sqrt{2x+y} \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} 2x+y \leq 25 \\ 5x=25-10\sqrt{2x+y} \end{matrix}\right.$

$\Rightarrow$ $\left\{\begin{matrix} 2x+y \leq 25 \\ 2\sqrt{2x+y}=5-x \end{matrix}\right.$
Thay $\sqrt{2x+y}=\frac{5-x}{2}$ vào $\sqrt{2x+y}+x-y=2 ^{(1)}$ ta được:
$\frac{5-x}{2}+x-y=2$

$\Leftrightarrow$ $5-x+2x-2y=4$
$\Leftrightarrow$ $x-y=-1$ $\Rightarrow$ Thế vào $(1)$ $\sqrt{2x+y}=3$
$\Rightarrow$ Ta có hệ: $\left\{\begin{matrix} x-y=-1 \\ 2x+y=9 \end{matrix}\right.$
$\Rightarrow$ $\left\{\begin{matrix} x=\frac{8}{3} \\ y=\frac{11}{3} \end{matrix}\right.$
Vậy hệ có 1 cặp nghiệm $(x;y)$ duy nhất là $(\frac{8}{3};\frac{11}{3})$

[BaNam]

10) $\left\{\begin{matrix} (4x^{2}+1)x+(y-3)\sqrt{5-2y}=0 \\ 4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right.$

ĐK: $y\leq\frac{5}{2}$ ; $x\leq \frac{3}{4}$
$\Rightarrow$ $\left\{\begin{matrix} 8x^{3}+2x-[(5-2y).\sqrt{5-2y}+\sqrt{5-2y}]=0 \\ 4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} (2x)^{3}-(\sqrt{5-2y})^{3}+(2x-\sqrt{5-2y})=0 \\ 4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right.$
$\Leftrightarrow$ $\left\{\begin{matrix} (2x-\sqrt{5-2y})[(2x)^{2}+2x\sqrt{5-2y}+(\sqrt{5-2y})^{2} +2]=0 \\ 4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right. $
$\Rightarrow$ $\left\{\begin{matrix} 2x=\sqrt{5-2y} \\  4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right.$

 

$\Rightarrow$ $\left\{\begin{matrix} 4x^{2}=5-2y \\ 4x^{2}+y^{2}+2\sqrt{3-4x}=7 \end{matrix}\right.$ 
$\Rightarrow$ $4x^{2}+(\frac{5-4x^{2}}{2})^{2}+2\sqrt{3-4x}=7$
$\Leftrightarrow$ $16x^{4}-24x^{2}-3+8\sqrt{3-4x}=0$ $^{(1)}$
$\Rightarrow$ $x=\frac{1}{2}$   (M chỉ biết có 1 nghiệm trên chứ chưa ra hướng giải)