Question

7√3 -5√2 upon √48+ √18.... simplify by rationalising the denominator

Answer 1

Answer:

$\frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{48}+\sqrt{18})}\\=\frac{114-41\sqrt{6}}{30}$

Step-by-step explanation:

$Given \: \frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{48}+\sqrt{18})}$

=$\frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{4\times 4\times 3}+\sqrt{3\times 3\times 2})}$

=$\frac{(7\sqrt{3}-5\sqrt{2})}{(4\sqrt{3}+3\sqrt{2})}$

/* Multiply numerator and denominator by $(4\sqrt{3}-3\sqrt{2})$, we get

=$\frac{(7\sqrt{3}-5\sqrt{2})(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}+3\sqrt{2})(4\sqrt{3}-3\sqrt{2})}$

=$\frac{28\times 3-21\sqrt{6}-20\sqrt{6}+15\times 2}{(4\sqrt{3})^{2}-(3\sqrt{2})^{2}}$

=$\frac{84-41\sqrt{6}+30}{48-18}$

=$\frac{114-41\sqrt{6}}{30}$

/* Denominator rationalised .

Therefore,

$\frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{48}+\sqrt{18})}\\=\frac{114-41\sqrt{6}}{30}$

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Answer 2

Answer:

114-41✓6/30

I hope I help your