Răspuns:
Explicație pas cu pas:
[tex]3a)~\frac{3}{2\sqrt{3} }=\frac{(\sqrt{3})^{2} }{2\sqrt{3}}=\frac{\sqrt{3} }{2}\\3b)~\frac{1}{2-\sqrt{3} }=\frac{2+\sqrt{3} }{(2-\sqrt{3})(2+\sqrt{3})}=\frac{2+\sqrt{3} }{2^{2}-(\sqrt{3})^{2}}=\frac{2+\sqrt{3} }{4-3} =2+\sqrt{3}\\3c)~a=\frac{3}{\sqrt{3} }=\frac{(\sqrt{3})^{2} }{\sqrt{3} }=\sqrt{3}\\[/tex]
[tex]b=\frac{2}{\sqrt{2} }=\frac{(\sqrt{2})^{2} }{\sqrt{2} }=\sqrt{2}\\\sqrt{x} \sqrt{3}>\sqrt{2}~deci~a>b\\[/tex]
[tex]\frac{\sqrt{2}-1 }{\sqrt{2} }+\frac{\sqrt{3}-\sqrt{2} }{\sqrt{6} } =\frac{\sqrt{2} }{\sqrt{2} }-\frac{1}{\sqrt{2} } +\frac{\sqrt{3} }{\sqrt{6} }-\frac{\sqrt{2} }{\sqrt{6} } =1-\frac{1}{\sqrt{2} } +\frac{1}{\sqrt{2} } -\frac{1}{\sqrt{3} } =1-\frac{1}{\sqrt{3} } =1-\frac{\sqrt{3} }{3 } =\frac{3-\sqrt{3} }{3}[/tex]
ex4a) (3x²-5x+7)-(2x²+7x-5)= 3x²-5x+7-2x²-7x+5=x²-12x+12.
b) 3x(9x²+3x-1)=27x³+9x²-3x
c) (x²√2-1)(x²√2+1)=(x²√2)²-1²=x⁴·(√2)²-1=2x⁴-1
