Rozwiąż zadanie z załącznika

[tex]a=3^{0,5}\cdot3^{-1,25}\cdot3^{0,75}=3^{0,5-1,25+0,75}=3^{1,25-1,25}=3^0=1\\\\ b=3^{0,5}\cdot18+3^{2,5}=3^{\frac{1}{2}}\cdot18+3^{\frac{25}{10}}=3^{\frac{1}{2}}\cdot18+3^{\frac{5}{2}}=18\cdot3^{\frac{1}{2}}+3^{\frac{5}{2}}=(18+3^2)\cdot3^{\frac{1}{2}}=\\\\=(18+9)\cdot3^{\frac{1}{2}}=27\cdot3^{\frac{1}{2}}=3^3\cdot3^{\frac{1}{2}}=3^{3+\frac{1}{2}}=3^{\frac{7}{2}}\\\\\\\dfrac{b}{a}=\dfrac{3^{\frac{7}{2}}}{1}=3^{\frac{7}{2}}\\\\\\Odp.C[/tex]

Marsuwviz Marsuwviz · Answer 2

Marsuwviz Marsuwviz

Odpowiedź:

[tex]b=3^{0,5}*18+3^{2,5}=3^\frac{1}{2} *3^2*2+3^{\frac{5}{2}}= 3^{\frac{5}{2}}*2+3^{\frac{5}{2}}=3^{\frac{5}{2}}(2+1)=3^{\frac{5}{2}}*3=3^{\frac{7}{2}}=\sqrt3^7=3^3*\sqrt3\\a=3^{0,5}*3^{-1,25}*3^{0,75} =3^{\frac{2}{4} }*3^{-\frac{5}{4}}*3^{\frac{3}{4}} =3^{(\frac{2}{4} -\frac{5}{4}+\frac{3}{4})}=3^0=1\\\frac{b}{a} =\frac{3^3*\sqrt3}{1} =3^3*\sqrt3[/tex]

Odp:C

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