Chủ đề này có 5 trả lời

[lovengan22]

GHPT:

a/ $1.\sqrt{x+1}+\sqrt{y}=\frac{13}{2}  $

     $2..\sqrt{x-2}+\sqrt{y-3}=\frac{11}{2}$

b/ 1. $2x^{2}y -xy=xy^{2}-2x+y$

    2. $(x^{2}+2y^{2})(1+\frac{1}{xy})^{2} = \frac{11}{2}$

Bài viết đã được chỉnh sửa nội dung bởi tpdtthltvp: 15-01-2017 - 08:22

[Master Kaiser]

GHPT:

a/ $1.\sqrt{x+1}+\sqrt{y}=\frac{13}{2}  $

     $2..\sqrt{x-2}+\sqrt{y-3}=\frac{11}{2}$

$hpt\Leftrightarrow \left\{\begin{matrix} 4(x+y+1+2\sqrt{y(x+1)})=169 & & \\ 4(x+y-5+2\sqrt{(x-2)(y-3)})=121 & & \end{matrix}\right.$

Trừ vế , ta được : $4(6+2(\sqrt{y(x+1)}-\sqrt{(x-2)(y-3)}))=48$

$\Leftrightarrow \sqrt{y(x+1)}=3+\sqrt{(x-2)(y-3)}$

$\Leftrightarrow xy+y=9+(x-2)(y-3)+6\sqrt{(x-2)(y-3)}$

$\Leftrightarrow x+y-5=2\sqrt{(x-2)(y-3)}$

$\Leftrightarrow (x+y-5)^2-4(x-2)(y-3)=0$

$\Leftrightarrow (x-y+1)^2=0$

$\Rightarrow y=x+1$

 

Thay vào (1) ... OK?

Bài viết đã được chỉnh sửa nội dung bởi Master Kaiser: 17-01-2017 - 18:29

[Master Kaiser]

GHPT:

a/ $1.\sqrt{x+1}+\sqrt{y}=\frac{13}{2}  $

     $2..\sqrt{x-2}+\sqrt{y-3}=\frac{11}{2}$

$hpt\Leftrightarrow \left\{\begin{matrix} 4(x+y+1+2\sqrt{y(x+1)})=169 & & \\ 4(x+y-5+2\sqrt{(x-2)(y-3)})=121 & & \end{matrix}\right.$

Trừ vế , ta được : $4(6+2(\sqrt{y(x+1)}-\sqrt{(x-2)(y-3)}))=48$

$\Leftrightarrow \sqrt{y(x+1)}=3+\sqrt{(x-2)(y-3)}$

$\Leftrightarrow xy+y=9+(x-2)(y-3)+6\sqrt{(x-2)(y-3)}$

$\Leftrightarrow x+y-5=2\sqrt{(x-2)(y-3)}$

$\Leftrightarrow (x+y-5)^2-4(x-2)(y-3)=0$

$\Leftrightarrow (x-y+1)^2=0$

$\Rightarrow y=x+1$

 

Thay vào (1) ... OK?

[Master Kaiser]

GHPT:

b/ 1. $2x^{2}y -xy=xy^{2}-2x+y$

    2. $(x^{2}+2y^{2})(1+\frac{1}{xy})^{2} = \frac{11}{2}$

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

$\Rightarrow 2(x^2+2y^2).\frac{x^2y^2}{(2x-y)^2}=11x^2y^2$

$\Leftrightarrow \frac{2(x^2+2y^2)}{(2x-y)^2}=11$

$\Leftrightarrow 2(x^2+2y^2)=11(2x-y)^2$

$\Leftrightarrow 42x^2+7y^2-44xy=0$

 

Xong nhé )

[Master Kaiser]

GHPT:

b/ 1. $2x^{2}y -xy=xy^{2}-2x+y$

    2. $(x^{2}+2y^{2})(1+\frac{1}{xy})^{2} = \frac{11}{2}$

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

$\Rightarrow 2(x^2+2y^2).\frac{x^2y^2}{(2x-y)^2}=11x^2y^2$

$\Leftrightarrow \frac{2(x^2+2y^2)}{(2x-y)^2}=11$

$\Leftrightarrow 2(x^2+2y^2)=11(2x-y)^2$

$\Leftrightarrow 42x^2+7y^2-44xy=0$

 

Xong nhé  )

[Master Kaiser]

Sr nhé mạng lag quá