b. 2;20;56;110;182;272;380

Quy luật là:1x2=2; 4x5=20; 7x8=56;10x11=110; 13x14=182; 16x17=272; 19x20=380

a. 2;12;30;56;90;132;182

quy luật là : 1x2=2; 3x4=12; 5x6=30; 7x8=56; 9x10=90; 11x12=132; 13x14=182

a. (x-1)x(x+1)(x+2)=24

[(x-1)(x+2)].[x(x+1)]=24

(\(x^2\)+2x-x-2)(\(x^2\)+x)=24

(\(x^2\)+x-2)(\(x^2\)+x)=24

[(\(x^2\)+x-1)-1].[(\(x^2\)+x-1)+1]=24

\(\left(x^2+x-1\right)^2\)-1=24

\(\left(x^2+x-1\right)^2\)=25

\(\left(x^2+x-1\right)^2\)=\(5^2\) hoặc\(\left(x^2+x-1\right)^2\)=\(\left(-5\right)^2\)

\(x^2\)+x-1=5 hoặc \(x^2\)+x-1=-5

\(x^2\)+x-6=0 hoặc \(x^2\)+x+4=0(vô nghiệm)

\(\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)

Vậy x=2 hoặc x=-3

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

B=\(\sqrt{\frac{1}{2}\left(6+2\sqrt{5}\right)}\)-\(\sqrt{\frac{1}{2}\left(6-2\sqrt{5}\right)}\)-\(\sqrt{2}\)

B=\(\sqrt{\frac{1}{2}\left(5+2\sqrt{5}.1+1\right)}\)-\(\sqrt{\frac{1}{2}\left(5-2\sqrt{5}.1+1\right)}\)-\(\sqrt{2}\)

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

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

B=\(\frac{2}{\sqrt{2}}\)-\(\sqrt{2}\)

B=\(\sqrt{2}\)-\(\sqrt{2}\)=0