Câu hỏi của Tuyet Thanh Tran

Question

tìm x, y biết:

$\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\sqrt{x}+\sqrt{y}$  =4

Nguyễn Danh Hoàng · Accepted Answer

$\dfrac{1}{\sqrt{x}}$$+$$\dfrac{1}{\sqrt{y}}$$+$$\sqrt{x}$$+$$\sqrt{y}$$=4$

$\Leftrightarrow$$\dfrac{1}{\sqrt{x}}$$-1$$\dfrac{1}{\sqrt{y}}$-1+$\sqrt{x}$-1$\sqrt{y}$-1=0

$\Leftrightarrow$$\dfrac{\sqrt{x}-1}{\sqrt{x}}+\sqrt{x}-1+\dfrac{\sqrt{y}-1}{\sqrt{y}}+\sqrt{y}-1$=0

$\Leftrightarrow$$\left(\sqrt{x}-1\right)\left(\dfrac{1}{\sqrt{x}}+1\right)$$+\left(\sqrt{y}-1\right)\left(\dfrac{1}{\sqrt{y}}+1\right)=0$

$\Leftrightarrow$$\left[{}\begin{matrix}\sqrt{y}=1\\sqrt{x}=1\end{matrix}\right.$

$\Leftrightarrow$$\left[{}\begin{matrix}y=1\x=1\end{matrix}\right.$

VậyCâu trả lời: $\dfrac{1}{\sqrt{x}}$$+$$\dfrac{1}{\sqrt{y}}$$+$$\sqrt{x}$$+$$\sqrt{y}$$=4$

$\Leftrightarrow$$\dfrac{1}{\sqrt{x}}$$-1$$\dfrac{1}{\sqrt{y}}$-1+$\sqrt{x}$-1$\sqrt{y}$-1=0

$\Leftrightarrow$$\dfrac{\sqrt{x}-1}{\sqrt{x}}+\sqrt{x}-1+\dfrac{\sqrt{y}-1}{\sqrt{y}}+\sqrt{y}-1$=0

$\Leftrightarrow$$\left(\sqrt{x}-1\right)\left(\dfrac{1}{\sqrt{x}}+1\right)$$+\left(\sqrt{y}-1\right)\left(\dfrac{1}{\sqrt{y}}+1\right)=0$

$\Leftrightarrow$$\left[{}\begin{matrix}\sqrt{y}=1\\sqrt{x}=1\end{matrix}\right.$

$\Leftrightarrow$$\left[{}\begin{matrix}y=1\x=1\end{matrix}\right.$

Vậy

Bài 8: Rút gọn biểu thức chứa căn bậc hai