Completely simplify and rationalize the denominator:$$\frac{\sqrt{160}}{\sqrt{252}}\times\frac{\sqrt{245}}{\sqrt{3}}$$
+9519 The method hipie gave is also possible, but it is tedious. Instead, we could simplify the square roots on their own first.
\(\quad \dfrac{\sqrt{160}}{\sqrt{252}} \times \dfrac{\sqrt{245}}{\sqrt 3}\\ =\dfrac{4 \sqrt{10}}{6 \sqrt 7} \times \dfrac{7 \sqrt 5}{\sqrt 3}\\ = \dfrac{4\times 7}{6} \times \sqrt{\dfrac{10 \times 5}{7 \times 3}}\\ = \dfrac{4\times 7 \times \sqrt{10 \times 5 \times 7 \times 3}}{6\times 7 \times 3}\\ = \dfrac{2\sqrt{1050}}{9}\\ = \dfrac{10 \sqrt{42}}{9}\)
