Q1) \(x + 1\over 2\) ÷ \(4 \over {x + 6}\) = [ \(x^2 + 7 x + 6\over 8\) ]

Q1) \(x + 8\over 1\) ÷ \({ x + 7} \over 2 \) = [ \(2( x + 8) \over( x + 7)\) ]

Q1) \(x + 9\over 4\) x \(x + 3\over x + 9\) = [ \(x + 3\over 4\) ]

Q2) \(x + 7\over 10\) ÷ \(5 \over {x + 4}\) = [ \(x^2 + 11x + 28\over 50\) ]

Q2) \(x + 5\over 2\) x \(7 \over{ x + 8}\) = [ \(7( x + 5) \over 2 ( x + 8)\) ]

Q2) \(x + 3\over 7\) ÷ \( x + 3\over x + 2\) = [ \(x + 2\over 7\) ]

Q3) \(x + 2\over 9\) x \(x + 4\over 6\) = [ \(x^2 + 6 x + 8\over 54\) ]

Q3) \(x + 2\over 5\) ÷ \({ x + 10} \over 6 \) = [ \(6( x + 2) \over 5 ( x + 10)\) ]

Q3) \(x + 5\over 9\) ÷ \( x + 5\over x + 8\) = [ \(x + 8\over 9\) ]

Q4) \(x + 3\over 5\) ÷ \(10 \over {x + 8}\) = [ \(x^2 + 11x + 24\over 50\) ]

Q4) \(x + 6\over 3\) x \(1 \over{ x + 5}\) = [ \(1( x + 6) \over 3 ( x + 5)\) ]

Q4) \(x + 4\over 4\) x \(x + 3\over x + 4\) = [ \(x + 3\over 4\) ]

Q5) \(x + 1\over 8\) x \(x + 5\over 2\) = [ \(x^2 + 6 x + 5\over 16\) ]

Q5) \(x + 3\over 5\) x \(2 \over{ x + 7}\) = [ \(2( x + 3) \over 5 ( x + 7)\) ]

Q5) \(x + 2\over 4\) x \(x + 5\over x + 2\) = [ \(x + 5\over 4\) ]

Q6) \(x + 6\over 7\) ÷ \(2 \over {x + 3}\) = [ \(x^2 + 9 x + 18\over 14\) ]

Q6) \(x + 1\over 5\) ÷ \({ x + 3} \over 3 \) = [ \(3( x + 1) \over 5 ( x + 3)\) ]

Q6) \(x + 7\over 7\) x \(x + 4\over x + 7\) = [ \(x + 4\over 7\) ]

Q7) \(x + 10\over 10\) ÷ \(5 \over {x + 9}\) = [ \(x^2 + 19 x + 90\over 50\) ]

Q7) \(x + 2\over 1\) ÷ \({ x + 5} \over 2 \) = [ \(2( x + 2) \over( x + 5)\) ]

Q7) \(x + 2\over 2\) x \(x + 2\over x + 2\) = [ \(x + 2\over 2\) ]

Q8) \(x + 1\over 9\) ÷ \(6 \over {x + 7}\) = [ \(x^2 + 8 x + 7\over 54\) ]

Q8) \(x + 6\over 3\) x \(1 \over{ x + 9}\) = [ \(1( x + 6) \over 3 ( x + 9)\) ]

Q8) \(x + 2\over 5\) x \(x + 4\over x + 2\) = [ \(x + 4\over 5\) ]

Q9) \(x + 8\over 7\) ÷ \(4 \over {x + 4}\) = [ \(x^2 + 12 x + 32\over 28\) ]

Q9) \(x + 1\over 5\) x \(2 \over{ x + 4}\) = [ \(2( x + 1) \over 5 ( x + 4)\) ]

Q9) \(x + 7\over 4\) ÷ \( x + 7\over x + 8\) = [ \(x + 8\over 4\) ]

Q10) \(x + 9\over 7\) x \(x + 8\over 5\) = [ \(x^2 + 17 x + 72\over 35\) ]

Q10) \(x + 6\over 2\) ÷ \({ x + 7} \over 3 \) = [ \(3( x + 6) \over 2 ( x + 7)\) ]

Q10) \(x + 4\over 6\) x \(x + 3\over x + 4\) = [ \(x + 3\over 6\) ]