Câu hỏi của Nguyễn Mạnh

Question

tim x bt

x²-(căn3 +căn5)*x+căn3 * căn5=0

Hoàng Phúc · Accepted Answer

$x^2-\left(\sqrt{3}+\sqrt{5}\right).x+\sqrt{3}.\sqrt{5}=0$

$\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0$

$\Leftrightarrow x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0$

$\Leftrightarrow\left(x-\sqrt{5}\right)\left(x-\sqrt{3}\right)=0$

$\Leftrightarrow\int^{x-\sqrt{5}=0}_{x-\sqrt{3}=0}\Leftrightarrow\int^{x=\sqrt{5}}_{x=\sqrt{3}}$

Vậy x $\in\left\{\sqrt{3};\sqrt{5}\right\}$

Câu trả lời:

$x^2-\left(\sqrt{3}+\sqrt{5}\right).x+\sqrt{3}.\sqrt{5}=0$

$\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0$

$\Leftrightarrow x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0$

$\Leftrightarrow\left(x-\sqrt{5}\right)\left(x-\sqrt{3}\right)=0$

$\Leftrightarrow\int^{x-\sqrt{5}=0}_{x-\sqrt{3}=0}\Leftrightarrow\int^{x=\sqrt{5}}_{x=\sqrt{3}}$

Vậy x $\in\left\{\sqrt{3};\sqrt{5}\right\}$