Question

\[\sqrt{48}-\sqrt{75}+\sqrt{192}\]

Answer 1

Try to factor each number:
48 = 2*2*2*2*3
75=3*5*5
192=2*2*2*2*2*2**3

Answer 2

\[\sqrt{2^4\cdot 3} - \sqrt{5^2 \cdot 3} +\sqrt{2^6\cdot3}\]

Answer 3

You can cancel numbers in pairs and take one out. For example:\[\sqrt{\cancel{2\cdot 2} \cdot 2\cdot 2\cdot 3} = 2\sqrt{\cancel{2\cdot 2} \cdot 3} = 2\cdot 2\sqrt3 = 4\sqrt3\]And so on...

Answer 4

Just look at @ParthKohli's tips!

Answer 5

Another thing is that you can add radicals. For example:\[4\sqrt 5 + 9 \sqrt 5 = (4 + 9)\sqrt 5 = 13\sqrt 5\]

Answer 6

so you can cross out all the 2's in 2^4

Answer 7

Yes, and write two \(2's\) outside.

Answer 8

\[ so its like 2\sqrt{6} i think \right\]

Answer 9

\[ Which ratio forms a proportion \frac{ 18 }{ 38 } with ?\]