Question

ma ajutați va rog???????

Answer 1

14)   
[tex]\displaystyle \\ \left(3^{ \frac{1}{2}} -2^{ \frac{1}{2} \right)\left(3^{ \frac{1}{2}} +2^{ \frac{1}{2} \right) = \left(3^{ \frac{1}{2}} \right)^2 - \left(2^{ \frac{1}{2}} \right)^2 =3^{ \frac{1}{2} \times 2} - 2^{ \frac{1}{2} \times 2} = 3 - 2 =\boxed{1} \\ \\ \\ \left(x^{ \frac{1}{2}} +y^{ \frac{1}{2} \right)\left(x^{ \frac{1}{2}} -y^{ \frac{1}{2} \right) = \left(x^{ \frac{1}{2}} \right)^2 - \left(y^{ \frac{1}{2}} \right)^2 =x^{ \frac{1}{2} \times 2} - y^{ \frac{1}{2} \times 2} =\boxed{x-y}[/tex]

15)
[tex]a) \\ \displaystyle \\ 2^{ \frac{2}{x} }= \sqrt[3]{2} \\ \\ 2^{ \frac{2}{x} }=2^{ \frac{1}{3} } \\ \\ \frac{2}{x} = \frac{1}{3} \\ \\ x = \frac{2 \cdot 3}{1} = \boxed{6} \\ \\ \\ c) \\ \sqrt{2}\cdot 2^{\frac{2}{x}}=\sqrt[3]{2}\\\\ 2^{\frac{1}{2}}\cdot2^{\frac{2}{x}}}=\sqrt[3]{2}\\\\ 2^{ \frac{1}{2} \cdot \frac{2}{x} } =2^{ \frac{1}{3}} \\ \\ \frac{1}{2} \cdot \frac{3}{x} = \frac{1}{3} \\ \\ \frac{3}{x} = \frac{2}{3} \\ \\ x = \frac{3 \cdot 3}{2} =\boxed{\frac{9}{2} }[/tex]



[tex]e) \\ \displaystyle \\ \Big( \sqrt{2} \cdot \sqrt[3]{2} \Big)^{ \frac{x+1}{3} }=\Big(\sqrt[3]{4} \Big)^{ \frac{1}{3} } \\ \\ \Big( 2^{ \frac{1}{2} } \cdot 2^{ \frac{1}{3} } \Big)^{ \frac{x+1}{3} }=\Big( 2^{2\cdot \frac{1}{3} } \Big)^{ \frac{1}{3} } \\ \\ \Big( \frac{1}{2} + \frac{1}{3} \Big) \cdot \frac{x+1}{3} = 2\cdot \frac{1}{3}\cdot \frac{1}{3} \\ \\ \frac{5}{6} \cdot \frac{x+1}{3} = 2\cdot \frac{1}{3}\cdot \frac{1}{3} \\ \\ \frac{5x+5}{18} = \frac{2}{9} [/tex]


[tex]\displaystyle \\ \frac{5x+5}{18} = \frac{2}{9} \\ \\ 5x+5 = \frac{2 \cdot 18}{9} \\ \\ 5x+5 = 4 \\ \\ 5x = 4 - 5 \\ \\ 5x= -1 \\ \\ x = \frac{-1}{5} = \boxed{-\frac{1}{5} }[/tex]