\(A=\frac{4\sqrt{x}}{\sqrt{x}+3}\left(\frac{x}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}+15}{x-9}\right)\)ĐK : \(x\ge0;x\ne9\)

\(=\frac{4\sqrt{x}}{\sqrt{x}+3}\left(\frac{x\sqrt{x}-3x+x+3\sqrt{x}+\sqrt{x}+15}{x-9}\right)\)

\(=\frac{4\sqrt{x}}{\sqrt{x}+3}\left(\frac{x\sqrt{x}-2x+4\sqrt{x}+15}{x-9}\right)\)

\(2a-2\sqrt{a}=2\sqrt{a}\left(\sqrt{a}-1\right)\)

Chúc học tốt 

ĐK \(x\le2\)Đặt \(\sqrt{2-x}=t\Rightarrow2-x=t^2\)\(\Rightarrow x=2-t^2\)ta có

\(N=2-t^2+t\)\(=-\left(t^2-2t\frac{1}{2}+\frac{1}{4}\right)+\frac{9}{4}=-\left(t-\frac{1}{2}\right)^2+\frac{9}{4}\ge\frac{9}{4}\)

Vì \(-\left(t-\frac{1}{2}\right)^2\le0\)

Dấu = xảy ra khi và chỉ khi \(t-\frac{1}{2}=0\Rightarrow t=\frac{1}{2}\Rightarrow\sqrt{2-x}=\frac{1}{2}\Leftrightarrow2-x=\frac{1}{4}\Rightarrow x=\frac{7}{4}\)tmđk

Vậy Max_N=9/4 <=> x=7/4

\(A=\sqrt{9x^2-12x+4}+1-3x=\sqrt{\left(3x-2\right)^2}+1-3x\)

\(=\left|3x-2\right|+1-3x\)Thay x = 1/3 vào A ta được : 

\(=\left|\frac{1}{3}.3-2\right|+1-\frac{3.1}{3}=1+1-1=1\)

-(5*căn bậc hai(2)*căn bậc hai(3)-17*căn bậc hai(2)-5)/(5*căn bậc hai(2))

\(\frac{2}{5-\sqrt{3}}+\frac{5}{2\sqrt{2}+3}-\sqrt{8}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{22}+\frac{5\left(2\sqrt{2}-3\right)}{-1}-\sqrt{8}\)

\(=\frac{5+\sqrt{3}}{11}-10\sqrt{2}+15-2\sqrt{2}\)

\(=\frac{5+\sqrt{3}}{11}-12\sqrt{2}+15=\frac{5+\sqrt{3}}{11}-\frac{11\left(12\sqrt{2}-15\right)}{11}\)

\(=\frac{5+\sqrt{3}-132\sqrt{2}+165}{11}=\frac{\sqrt{3}-132\sqrt{2}+170}{11}\)

\(\frac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3-2\sqrt{6}+2}}{\sqrt{3}-\sqrt{2}}\)

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

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\(\sqrt{2x+8}=0\)ĐK : \(x\ge-4\)

\(\Leftrightarrow2x+8=0\Leftrightarrow x=-4\)