Lời giải:

a) Đặt biểu thức là $A$.

\(A=\frac{(1-\sqrt{2})^2-(1+\sqrt{2})^2}{(1-\sqrt{2})(1+\sqrt{2})}.\frac{1}{6\sqrt{2}}=\frac{-4\sqrt{2}}{-1}.\frac{1}{6\sqrt{2}}=\frac{2}{3}\)

b) Để hàm số $y=(\sqrt{m}-2)x+3$ đồng biến thì $\sqrt{m}-2>0$

$\Leftrightarrow m>4$

a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

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

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

\(=-4-2\sqrt{6}\)

Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

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

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).

\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=\dfrac{\left(\sqrt{2-\sqrt{3}}\right)^2+\left(\sqrt{2+\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}=\dfrac{4}{1}=4\)

a, \(A=2\sqrt{3}-\sqrt{12}-\sqrt{9}\)

\(=2\sqrt{3}-2\sqrt{3}-3=-3\)

b, \(B=\sqrt{3}\left(\sqrt{12}+\sqrt{27}\right)\)

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

\(=\sqrt{3}.5\sqrt{3}=5.3=15\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)

Câu 1: 

1: Ta có: \(16\sqrt{9}-9\sqrt{16}\)

\(=16\cdot3-9\cdot4\)

\(=48-36=12\)

2:

a) Thay x=2 và y=8 vào hàm số \(y=a\cdot x^2\), ta được:

\(a\cdot2^2=8\)

\(\Leftrightarrow4a=8\)

hay a=2

Vậy: a=2

Câu 2: 

1: 

a) Ta có: \(x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: S={1;4}