Q1) \(x + 2\over 8\) ÷ \(6 \over {x + 4}\) = [ \(x^2 + 6 x + 8\over 48\) ]

Q1) \(x + 10\over 1\) ÷ \({ x + 4} \over 3 \) = [ \(3( x + 10) \over( x + 4)\) ]

Q1) \(x + 4\over 5\) ÷ \( x + 4\over x + 7\) = [ \(x + 7\over 5\) ]

Q2) \(x + 10\over 10\) x \(x + 6\over 2\) = [ \(x^2 + 16 x + 60\over 20\) ]

Q2) \(x + 5\over 9\) ÷ \({ x + 9} \over 10 \) = [ \(10( x + 5) \over 9 ( x + 9)\) ]

Q2) \(x + 6\over 4\) x \(x + 8\over x + 6\) = [ \(x + 8\over 4\) ]

Q3) \(x + 6\over 9\) ÷ \(3 \over {x + 5}\) = [ \(x^2 + 11x + 30\over 27\) ]

Q3) \(x + 9\over 6\) ÷ \({ x + 2} \over 5 \) = [ \(5( x + 9) \over 6 ( x + 2)\) ]

Q3) \(x + 8\over 5\) ÷ \( x + 8\over x + 6\) = [ \(x + 6\over 5\) ]

Q4) \(x + 6\over 2\) x \(x + 9\over 7\) = [ \(x^2 + 15 x + 54\over 14\) ]

Q4) \(x + 8\over 2\) x \(1 \over{ x + 10}\) = [ \(1( x + 8) \over 2 ( x + 10)\) ]

Q4) \(x + 9\over 4\) ÷ \( x + 9\over x + 2\) = [ \(x + 2\over 4\) ]

Q5) \(x + 5\over 3\) x \(x + 8\over 10\) = [ \(x^2 + 13 x + 40\over 30\) ]

Q5) \(x + 10\over 5\) ÷ \({ x + 7} \over 4 \) = [ \(4( x + 10) \over 5 ( x + 7)\) ]

Q5) \(x + 4\over 7\) x \(x + 6\over x + 4\) = [ \(x + 6\over 7\) ]

Q6) \(x + 7\over 2\) ÷ \(2 \over {x + 8}\) = [ \(x^2 + 15 x + 56\over 4\) ]

Q6) \(x + 4\over 2\) ÷ \({ x + 1} \over 1 \) = [ \(1( x + 4) \over 2 ( x + 1)\) ]

Q6) \(x + 3\over 7\) ÷ \( x + 3\over x + 6\) = [ \(x + 6\over 7\) ]

Q7) \(x + 7\over 6\) ÷ \(4 \over {x + 4}\) = [ \(x^2 + 11x + 28\over 24\) ]

Q7) \(x + 2\over 3\) ÷ \({ x + 4} \over 10 \) = [ \(10( x + 2) \over 3 ( x + 4)\) ]

Q7) \(x + 4\over 6\) x \(x + 5\over x + 4\) = [ \(x + 5\over 6\) ]

Q8) \(x + 7\over 8\) x \(x + 9\over 8\) = [ \(x^2 + 16 x + 63\over 64\) ]

Q8) \(x + 2\over 8\) x \(5 \over{ x + 7}\) = [ \(5( x + 2) \over 8 ( x + 7)\) ]

Q8) \(x + 4\over 10\) x \(x + 5\over x + 4\) = [ \(x + 5\over 10\) ]

Q9) \(x + 2\over 2\) ÷ \(7 \over {x + 7}\) = [ \(x^2 + 9 x + 14\over 14\) ]

Q9) \(x + 3\over 4\) ÷ \({ x + 9} \over 5 \) = [ \(5( x + 3) \over 4 ( x + 9)\) ]

Q9) \(x + 3\over 7\) ÷ \( x + 3\over x + 8\) = [ \(x + 8\over 7\) ]

Q10) \(x + 9\over 3\) ÷ \(9 \over {x + 8}\) = [ \(x^2 + 17 x + 72\over 27\) ]

Q10) \(x + 7\over 3\) ÷ \({ x + 5} \over 1 \) = [ \(1( x + 7) \over 3 ( x + 5)\) ]

Q10) \(x + 1\over 8\) x \(x + 10\over x + 1\) = [ \(x + 10\over 8\) ]