tolong ajarin iniiiiii​

Penyelesaiana. [tex](\frac{2}{4\sqrt{b^3}})[/tex][tex]4\sqrt{b^3} = 4b^{3/2}[/tex][tex]\frac{2}{4\sqrt{b^3}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{2\sqrt{b}}{4\sqrt{b^3}\sqrt{b}}[/tex][tex]\sqrt{b^3}\sqrt{b} = \sqrt{b^{3+1}} = \sqrt{b^4}[/tex][tex]= b^2[/tex][tex]=\frac{2\sqrt{b}}{4b^2}[/tex][tex]\frac{2\sqrt{b}}{4b^2} = \frac{\sqrt{b}}{2b^2}[/tex]b. [tex](\frac{2}{\sqrt{3}+\sqrt{5}})[/tex][tex]\frac{2}{\sqrt{3}+\sqrt{5}} \times \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} = \frac{2(\sqrt{3}-\sqrt{5})}{(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5})}[/tex][tex](\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5}) = (\sqrt{3})^2 - (\sqrt{5})^2[/tex][tex]= 3 - 5[/tex][tex]= -2[/tex][tex]\frac{2(\sqrt{3}-\sqrt{5})}{-2} = -(\sqrt{3}-\sqrt{5})[/tex][tex]= \sqrt{5} - \sqrt{3}[/tex]c. [tex](\frac{m}{\sqrt{m}+n})[/tex][tex]\frac{m}{\sqrt{m}+n} \times \frac{\sqrt{m}-n}{\sqrt{m}-n} = \frac{m(\sqrt{m}-n)}{(\sqrt{m}+n)(\sqrt{m}-n)}[/tex][tex](\sqrt{m}+n)(\sqrt{m}-n) = (\sqrt{m})^2 - n^2[/tex][tex]= m - n^2[/tex][tex]=\frac{m(\sqrt{m}-n)}{m-n^2}[/tex]

Answered by vinganzbeut | 2025-08-04

Bantu kasih like-nya ya apabila jawaban ini membantu. Terima kasih.

Answered by psycount | 2025-08-04