[tex] \frac{ \sqrt{3} }{2 \sqrt{2} + 3} = [/tex]Can You Solve This?.... Question from @Entrity

LyraeChan

Penjelasan dengan langkah-langkah:

Yes, I can.

The way to solve the square root is to multiply the denominator of fraction with the surd.

$= \frac{ \sqrt{3} }{2 \sqrt{2} + 3 }$

$= \frac{ \sqrt{3} }{2 \sqrt{2} + 3 } \times \frac{2 \sqrt{2} - 3 }{2 \sqrt{2} - 3}$

$= \frac{ \sqrt{3}.(2 \sqrt{2} - 3) }{ {(2 \sqrt{2} )}^{2} - {3}^{2} }$

$= \frac{2 \sqrt{6} - 3 \sqrt{3} }{8 - 9}$

$= \frac{2 \sqrt{6} - 3 \sqrt{3} }{ - 1}$

$= - 2 \sqrt{6} + 3 \sqrt{3} \: \text{atau} \: 3\sqrt{3} - 2\sqrt{6}$

2 votes Thanks 2

Entrity This is not true, the negative square root can be removed and converted to a positive number
so the correct answer is 3√3 - 2√6

LyraeChan loh tinggal dibolak balik kan jadi :D

LyraeChan dari -2√6 + 3√3 jadi 3√3 - 2√6

LyraeChan "this is not true" darimana :')

LyraeChan emang aturannya kyk gitu yah? maaf aku baru tau

xcvi

Jawab:
= 3√3 - 2√6

Penjelasan dengan langkah-langkah:
Rasionalisasi pecahan (a+b)(a-b) = a²-b²

$\frac{\sqrt{3}}{2\sqrt{2}+3}=\frac{\sqrt{3}}{2\sqrt{2}+3}\times\frac{2\sqrt{2}-3}{2\sqrt{2}-3}\\=\frac{\sqrt{3}(2\sqrt{2}-3)}{(2\sqrt{2})^2-3^2}=\frac{2\sqrt{6}-3\sqrt{3}}{2^2\cdot2-9}\\=\frac{2\sqrt{6}-3\sqrt{3}}{4\cdot2-9}=\frac{2\sqrt{6}-3\sqrt{3}}{8-9}$
= -(2√6 - 3√3)
= -(-3√3 + 2√6)
= 3√3 - 2√6

(Xcvi)

2 votes Thanks 1

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