a: Để (d)//y=x-3 thì \(4-m^2=1\)

\(\Leftrightarrow m\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

\(K=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=8-2\sqrt{15}\)

\(M=\sqrt{\dfrac{\left(11+\sqrt{96}\right)^2}{121-96}}+\sqrt{\dfrac{\left(11-\sqrt{96}\right)^2}{121-96}}\\ M=\dfrac{11+\sqrt{96}}{5}+\dfrac{11-\sqrt{96}}{5}=\dfrac{22}{5}\)

\(D=\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{2-\sqrt{3}}-\dfrac{6}{3-\sqrt{3}}\)

\(=\sqrt{3}+1-6-3\sqrt{3}-3-\sqrt{3}\)

\(=-8-3\sqrt{3}\)

11.A 12.D 13.B 14.C 15.D 16.A

ADBCDA nha

x=4

Với x\(\ge1\)\(x-1-\sqrt{x-1}=0< =>x-1=\sqrt{x-1}< =>\left(x-1\right)^2=x-1< =>\left(x-1\right)^2-\left(x-1\right)=0< =>\left(x-1\right)\left(x-1-1\right)=0< =>\left(x-1\right)\left(x-2\right)=0\)\(< =>\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=1\left(TM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

\(\Leftrightarrow x-\sqrt{x}-2=0\left(1\right)\\Đặt:\sqrt{x}=t\left(x\ge0\right)\\ \left(1\right)\Leftrightarrow t^2-t-2=0\\ \Leftrightarrow t^2-2t+\left(t-2\right)=0\\ \Leftrightarrow t.\left(t-2\right)+\left(t-2\right)=0\\ \Leftrightarrow\left(t+1\right).\left(t-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t+1=0\\t-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}t=-1\left(Loại\right)\\t=2\left(Nhận\right)\end{matrix}\right.\\ Với:t=2\Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow x=4\)