Bonsoir,
On sait que [tex]\sqrt{ab} =\sqrt{a}* \sqrt{b}[/tex].
4) A = √162 = √(2*81) = √2 * √81 = 9√2
B = 3√2 * 5√14 = 3*5 * √(2*14) = 15√(2*2*7) = 15√(4*7) = 15 * √4 * √7 = 15*2*√7 = 30√7
C = 2√12 - 4√75 + 2√27
= 2√(3*4) - 4√(25*3) + 2√(9*3)
= 2*2√3 - 4*5√3 + 2*3√3
= 4√3 - 20√3 + 6√3
= - 10√3
5)
[tex]A = \frac{6}{\sqrt{3} } =\frac{6\sqrt{3} }{\sqrt{3} *\sqrt{3} }=\frac{6\sqrt{3} }{3} =2\sqrt{3}[/tex]
[tex]B = \frac{1}{2-\sqrt{2} } =\frac{2+\sqrt{2} }{(2-\sqrt{2})(2+\sqrt{2}) } =\frac{2+\sqrt{2} }{2^{2}-(\sqrt{2})^{2} } =\frac{2+\sqrt{2} }{4-2} =\frac{2+\sqrt{2} }{2} =1+\frac{\sqrt{2} }{2}[/tex]
[tex]C=\frac{\sqrt{20} }{2-\sqrt{5} } =\frac{\sqrt{4*5} }{2-\sqrt{5} } =\frac{2\sqrt{5} }{2-\sqrt{5} } =\frac{2\sqrt{5}(2+\sqrt{5}) }{(2-\sqrt{5})(2+\sqrt{5}) } =\frac{4\sqrt{5}+2*5 }{4-5} =\frac{4\sqrt{5}+10 }{-1} =-10-4\sqrt{5}[/tex]
