[tex]\begin{aligned} \frac{a \sqrt{a} + b \sqrt{b}}{\sqrt{a} + \sqrt{b}} &= \frac{(a \sqrt{a} + b \sqrt{b})(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} \\ &= \frac{a^{\frac{3}{2}} - ab^{\frac{1}{2}} + ba^{\frac{1}{2}} - b^{\frac{3}{2}}}{a - b} \\ &= \frac{a^{\frac{3}{2}} - b^{\frac{3}{2}} + \sqrt{ab}(a-b)}{a-b} \\ &= \frac{a^{\frac{3}{2}} - b^{\frac{3}{2}}}{a-b} + \sqrt{ab}. \end{aligned}[/tex]
DIKETAHUI
[tex]\sf \dfrac{a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } [/tex]
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DITANYA
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JAWAB
Cara 1
[tex]\begin{aligned}\sf \dfrac{a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } &=\sf \dfrac{ a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } \times \dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}}\\&=\sf \dfrac{(a\sqrt{a}+b\sqrt{b})(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} \\&=\sf \dfrac{ {a}^{2} - a \sqrt{ab} + b \sqrt{ab} - {b}^{2} }{a - b} \\&=\sf \dfrac{ {a}^{2} - {b}^{2} - a \sqrt{ab} + b \sqrt{ab} }{a - b}\\&=\sf \dfrac{(a + b)(a - b) - (a - b )\sqrt{ab} }{(a - b)} \\&=\sf \dfrac{(a + b)(\bcancel{a - b})}{(\bcancel{a - b})}- \dfrac{(\bcancel{a - b})\sqrt{ab} }{(\bcancel{a - b})} \\&=\sf a + b - \sqrt{ab} \end{aligned}[/tex]
Cara 2
[tex]\begin{aligned}\sf \dfrac{a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } &=\sf \dfrac{\bcancel{( \sqrt{a} + \sqrt{b})}(a - \sqrt{ab} + b)}{\bcancel{(\sqrt{a} + \sqrt{b})}} \\&=\sf a - \sqrt{ab} + b \\&=\sf a + b- \sqrt{ab} \end{aligned}[/tex]
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[tex]\begin{aligned} \frac{a \sqrt{a} + b \sqrt{b}}{\sqrt{a} + \sqrt{b}} &= \frac{(a \sqrt{a} + b \sqrt{b})(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} \\ &= \frac{a^{\frac{3}{2}} - ab^{\frac{1}{2}} + ba^{\frac{1}{2}} - b^{\frac{3}{2}}}{a - b} \\ &= \frac{a^{\frac{3}{2}} - b^{\frac{3}{2}} + \sqrt{ab}(a-b)}{a-b} \\ &= \frac{a^{\frac{3}{2}} - b^{\frac{3}{2}}}{a-b} + \sqrt{ab}. \end{aligned}[/tex]
Verified answer
DIKETAHUI
[tex]\sf \dfrac{a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } [/tex]
ㅤ
DITANYA
Bentuk sederhananya
ㅤ
JAWAB
Cara 1
[tex]\begin{aligned}\sf \dfrac{a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } &=\sf \dfrac{ a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } \times \dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}}\\&=\sf \dfrac{(a\sqrt{a}+b\sqrt{b})(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} \\&=\sf \dfrac{ {a}^{2} - a \sqrt{ab} + b \sqrt{ab} - {b}^{2} }{a - b} \\&=\sf \dfrac{ {a}^{2} - {b}^{2} - a \sqrt{ab} + b \sqrt{ab} }{a - b}\\&=\sf \dfrac{(a + b)(a - b) - (a - b )\sqrt{ab} }{(a - b)} \\&=\sf \dfrac{(a + b)(\bcancel{a - b})}{(\bcancel{a - b})}- \dfrac{(\bcancel{a - b})\sqrt{ab} }{(\bcancel{a - b})} \\&=\sf a + b - \sqrt{ab} \end{aligned}[/tex]
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Cara 2
[tex]\begin{aligned}\sf \dfrac{a\sqrt{a}+b\sqrt{b}}{ \sqrt{a} + \sqrt{b} } &=\sf \dfrac{\bcancel{( \sqrt{a} + \sqrt{b})}(a - \sqrt{ab} + b)}{\bcancel{(\sqrt{a} + \sqrt{b})}} \\&=\sf a - \sqrt{ab} + b \\&=\sf a + b- \sqrt{ab} \end{aligned}[/tex]