Q1) \(x + 8\over 6\) x \(x + 4\over 8\) = [ \(x^2 + 12 x + 32\over 48\) ]

Q1) \(x + 6\over 1\) x \(1 \over{ x + 5}\) = [ \(1( x + 6) \over( x + 5)\) ]

Q1) \(x + 3\over 6\) x \(x + 4\over x + 3\) = [ \(x + 4\over 6\) ]

Q2) \(x + 2\over 6\) x \(x + 9\over 6\) = [ \(x^2 + 11x + 18\over 36\) ]

Q2) \(x + 10\over 1\) x \(1 \over{ x + 4}\) = [ \(1( x + 10) \over( x + 4)\) ]

Q2) \(x + 8\over 8\) x \(x + 1\over x + 8\) = [ \(x + 1\over 8\) ]

Q3) \(x + 3\over 4\) x \(x + 1\over 3\) = [ \(x^2 + 4 x + 3\over 12\) ]

Q3) \(x + 9\over 7\) x \(3 \over{ x + 7}\) = [ \(3( x + 9) \over 7 ( x + 7)\) ]

Q3) \(x + 1\over 5\) ÷ \( x + 1\over x + 10\) = [ \(x + 10\over 5\) ]

Q4) \(x + 8\over 4\) x \(x + 3\over 7\) = [ \(x^2 + 11x + 24\over 28\) ]

Q4) \(x + 8\over 2\) ÷ \({ x + 7} \over 3 \) = [ \(3( x + 8) \over 2 ( x + 7)\) ]

Q4) \(x + 5\over 6\) ÷ \( x + 5\over x + 9\) = [ \(x + 9\over 6\) ]

Q5) \(x + 9\over 8\) ÷ \(2 \over {x + 7}\) = [ \(x^2 + 16 x + 63\over 16\) ]

Q5) \(x + 10\over 4\) ÷ \({ x + 8} \over 5 \) = [ \(5( x + 10) \over 4 ( x + 8)\) ]

Q5) \(x + 6\over 9\) ÷ \( x + 6\over x + 6\) = [ \(x + 6\over 9\) ]

Q6) \(x + 8\over 6\) ÷ \(8 \over {x + 2}\) = [ \(x^2 + 10 x + 16\over 48\) ]

Q6) \(x + 5\over 5\) ÷ \({ x + 1} \over 9 \) = [ \(9( x + 5) \over 5 ( x + 1)\) ]

Q6) \(x + 5\over 5\) x \(x + 7\over x + 5\) = [ \(x + 7\over 5\) ]

Q7) \(x + 6\over 2\) x \(x + 2\over 7\) = [ \(x^2 + 8 x + 12\over 14\) ]

Q7) \(x + 2\over 3\) x \(5 \over{ x + 5}\) = [ \(5( x + 2) \over 3 ( x + 5)\) ]

Q7) \(x + 2\over 8\) ÷ \( x + 2\over x + 1\) = [ \(x + 1\over 8\) ]

Q8) \(x + 10\over 10\) ÷ \(9 \over {x + 7}\) = [ \(x^2 + 17 x + 70\over 90\) ]

Q8) \(x + 5\over 5\) x \(3 \over{ x + 9}\) = [ \(3( x + 5) \over 5 ( x + 9)\) ]

Q8) \(x + 2\over 4\) x \(x + 8\over x + 2\) = [ \(x + 8\over 4\) ]

Q9) \(x + 8\over 2\) x \(x + 9\over 10\) = [ \(x^2 + 17 x + 72\over 20\) ]

Q9) \(x + 4\over 1\) ÷ \({ x + 7} \over 1 \) = [ \(1( x + 4) \over( x + 7)\) ]

Q9) \(x + 4\over 3\) ÷ \( x + 4\over x + 8\) = [ \(x + 8\over 3\) ]

Q10) \(x + 2\over 10\) x \(x + 7\over 4\) = [ \(x^2 + 9 x + 14\over 40\) ]

Q10) \(x + 2\over 1\) ÷ \({ x + 9} \over 1 \) = [ \(1( x + 2) \over( x + 9)\) ]

Q10) \(x + 1\over 4\) x \(x + 4\over x + 1\) = [ \(x + 4\over 4\) ]