địt ko em

địt ko em

Với x > 0 ; x \(\ne\)9 

a, \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}+\frac{2}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\left(\frac{3\sqrt{x}+1+2\left(\sqrt{x}-3\right)}{x-3\sqrt{x}}\right)\)

\(=\left(\frac{-3\sqrt{x}-9}{x-9}\right):\left(\frac{5\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)=\frac{-3}{\sqrt{x}-3}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}=\frac{-3\sqrt{x}}{5\left(\sqrt{x}-1\right)}\)

b, Ta có : \(B< 0\Rightarrow\frac{-3\sqrt{x}}{5\left(\sqrt{x}-1\right)}< 0\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

Kết hợp vói đk vậy x > 1 ; x \(\ne\)9

\(A=3\sqrt{3}-8\sqrt{3}+15\sqrt{3}=10\sqrt{3}\)

\(B=\left|\sqrt{5}-2\right|-\left|\sqrt{5}-3\right|=\sqrt{5}-2-\left(\sqrt{5}-3\right)=\sqrt{5}-2-\sqrt{5}+3=1\)

\(C=\left|2\sqrt{3}+1\right|+\left|2\sqrt{3}-5\right|=2\sqrt{3}+1+5-2\sqrt{3}=6\)

\(D=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}-3\right|=\sqrt{5}-2-\sqrt{5}+3=1\)

\(E=\frac{4\left(\sqrt{5}+2\right)}{5-4}-\frac{32\left(\sqrt{5}-1\right)}{5-1}=4\sqrt{5}+8-\frac{32\sqrt{5}-32}{4}=4\sqrt{5}+8-8\sqrt{5}+8=16-4\sqrt{5}\)

\(M=\frac{10\left(3\sqrt{2}+4\right)}{18-16}+\frac{28\left(3\sqrt{2}-2\right)}{18-4}=\frac{30\sqrt{2}+40}{2}+\frac{84\sqrt{2}-56}{14}=15\sqrt{2}+20+6\sqrt{2}-4=16+21\sqrt{2}\)

hỏi văn hay hỏi toán z

????????????