\(\frac{2x}{x-3}\sqrt{9\left(x-3\right)^2}-2x=\frac{2x}{x-3}\cdot3\left(3-x\right)-2x=-6x-2x=-8x\)(vì x < 0)

\(=\frac{2x}{x-3}\cdot3\left(3-x\right)-2x=\frac{-6x\left(x-3\right)}{x-3}-2x=-6x-2x=-8x\)

mn tìm đkxđ và rút gọn hộ e vs ạ

đề có sai sót khong zạy bạn =))

Với x >= 1 A = x - ( x - 1 ) = x - x + 1 = 1

Với x \(\ge1\)biểu thức A có dạng 

\\(A=x-\left(x-1\right)=x-x+1=1\)

\(1,\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\)

\(\frac{\sqrt{4+2\sqrt{3}}}{2}=\frac{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}{2}\)

\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}\)

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

\(2,\sqrt{3-\sqrt{5}}:\sqrt{2}\)

\(\frac{\sqrt{6-2\sqrt{5}}}{2}=\frac{\sqrt{\sqrt{5}^2-2\sqrt{5}+1}}{2}\)

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

\(\frac{\sqrt{5}-1}{2}\)

\(3,\frac{\sqrt{6+2\sqrt{5}}}{1+\sqrt{5}}\)

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

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

\(4,\sqrt{27+7\sqrt{5}}:\sqrt{2}\)

\(\frac{\sqrt{54+2.7\sqrt{5}}}{2}\)

\(\frac{\sqrt{7^2+2.7.\sqrt{5}+5}}{2}\)

\(\frac{\sqrt{\left(7+\sqrt{5}\right)^2}}{2}\)

\(=\frac{7+\sqrt{5}}{2}\)

3x + a + 1 = 0

=> 3x + a = -1

=> 3x = -1 - a

=> x = \(\frac{-1 - a}{3}\)

\(C=\left(\frac{a+\sqrt{a}}{a-1}-\frac{\sqrt{a}}{a-1}\right):\frac{\sqrt{a}+1}{a-1}\)

\(C=\frac{a}{a-1}\cdot\frac{a-1}{\sqrt{a}+1}\)

\(C=\frac{a}{\sqrt{a}+1}\)

Vậy ...

search google là xong mà chị