Question

a)Tìm ĐKXĐ.

b)Rút gọn.

N=\(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\times\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\).P=\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\times\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

Toán 9 ấy bài 8 SGK tập 1

Help me !!!! 

 

 

Answer 1

N=\(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right).\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\)

ĐKXĐ \(\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\3-x\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\x\ne3\\x\ge0\end{cases}}\)

\(=\left[\frac{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right].\frac{\sqrt{x}+\sqrt{3}}{3-x}\)

\(=\left(\sqrt{x}+\sqrt{3}-2\sqrt{x}\right).\frac{\sqrt{x}+\sqrt{3}}{3-x}\)

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

Answer 2

câu 2 ra |a-b| nha bn mik đăng rồi nhưng bị lỗi nên nó ko hiện lên