\(\frac{x-\sqrt{x}}{x+2\sqrt{x}+1}=\frac{1}{3}\)

\(\Leftrightarrow3\left(x-\sqrt{x}\right)=x+2\sqrt{x}+1\)

\(\Leftrightarrow3x-3\sqrt{x}=x+2\sqrt{x}+1\)

\(\Leftrightarrow2x-5\sqrt{x}-1=0\)

Đặt \(t=\sqrt{x}\)  ( đk: \(t\ge0\) )  phương trình trở thành: 

\(2t^2-5t-1=0\)

\(\Delta=\left(-5\right)^2-4.2.\left(-1\right)=33>0\)

\(t_1=\frac{5+\sqrt{33}}{2.2}=\frac{5+\sqrt{33}}{4}\left(N\right)\)

\(t_2=\frac{5-\sqrt{33}}{2.2}=\frac{5-\sqrt{33}}{4}\left(L\right)\)

* Ta có \(t=\frac{5+\sqrt{33}}{4}\Leftrightarrow\sqrt{x}=\frac{5+\sqrt{33}}{4}\Leftrightarrow x=\left(\frac{5+\sqrt{33}}{4}\right)^2=\frac{29+5\sqrt{33}}{8}\)

Vậy \(x=\frac{29+5\sqrt{33}}{8}\)

Học tốt !!! :) 

Nhà nghèo quá ko có tiền mua bông hút, lấy "nùi giẻ" lau là được rồi ! =)) 

\(=\sqrt{3+2+2.\sqrt{3}.\sqrt{2}}\)

\(=\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\sqrt{3}+\sqrt{2}\)

(1+\(\sqrt{5}\))\(^2\)

\(\frac{\sqrt{x^2y^2}}{xy}+\frac{\sqrt{\left(x-y\right)^2x^2}}{x\left(x-y\right)}-\frac{\sqrt{\left(x-y\right)y^2}}{y\left(x-y\right)}\)

\(=\sqrt{\frac{x^2y^2}{x^2y^2}}+\sqrt{\frac{\left(x-y\right)^2x^2}{x^2\left(x-y\right)^2}}-\sqrt{\frac{\left(x-y\right)y^2}{y^2\left(x-y\right)^2}}\)

\(=1+1-\frac{1}{\sqrt{x-y}}\)

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

x-y<0 thì sao ạ

\(\sqrt{5+2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{5}+1-1}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2-1}\)

\(=\sqrt{5}+1-1\)

\(=\sqrt{5}\)

Chieu dai 11 m

Chieu rong 7 m

chiều dài là    : 11 m

chiều rộng là :  7 m