Q1) \(x + 5\over 3\) x \(x + 8\over 9\) = [ \(x^2 + 13 x + 40\over 27\) ]

Q1) \(x + 8\over 2\) ÷ \({ x + 9} \over 3 \) = [ \(3( x + 8) \over 2 ( x + 9)\) ]

Q1) \(x + 8\over 6\) x \(x + 6\over x + 8\) = [ \(x + 6\over 6\) ]

Q2) \(x + 4\over 6\) ÷ \(5 \over {x + 3}\) = [ \(x^2 + 7 x + 12\over 30\) ]

Q2) \(x + 6\over 8\) x \(5 \over{ x + 8}\) = [ \(5( x + 6) \over 8 ( x + 8)\) ]

Q2) \(x + 7\over 5\) x \(x + 9\over x + 7\) = [ \(x + 9\over 5\) ]

Q3) \(x + 5\over 5\) x \(x + 8\over 9\) = [ \(x^2 + 13 x + 40\over 45\) ]

Q3) \(x + 10\over 9\) x \(4 \over{ x + 8}\) = [ \(4( x + 10) \over 9 ( x + 8)\) ]

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

Q4) \(x + 9\over 7\) ÷ \(9 \over {x + 8}\) = [ \(x^2 + 17 x + 72\over 63\) ]

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

Q4) \(x + 2\over 6\) ÷ \( x + 2\over x + 1\) = [ \(x + 1\over 6\) ]

Q5) \(x + 10\over 4\) x \(x + 4\over 10\) = [ \(x^2 + 14 x + 40\over 40\) ]

Q5) \(x + 7\over 10\) ÷ \({ x + 4} \over 3 \) = [ \(3( x + 7) \over 10 ( x + 4)\) ]

Q5) \(x + 2\over 9\) ÷ \( x + 2\over x + 7\) = [ \(x + 7\over 9\) ]

Q6) \(x + 8\over 7\) x \(x + 7\over 4\) = [ \(x^2 + 15 x + 56\over 28\) ]

Q6) \(x + 5\over 4\) x \(9 \over{ x + 10}\) = [ \(9( x + 5) \over 4 ( x + 10)\) ]

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

Q7) \(x + 3\over 5\) x \(x + 7\over 3\) = [ \(x^2 + 10 x + 21\over 15\) ]

Q7) \(x + 9\over 5\) ÷ \({ x + 8} \over 1 \) = [ \(1( x + 9) \over 5 ( x + 8)\) ]

Q7) \(x + 4\over 4\) ÷ \( x + 4\over x + 8\) = [ \(x + 8\over 4\) ]

Q8) \(x + 1\over 8\) x \(x + 9\over 8\) = [ \(x^2 + 10 x + 9\over 64\) ]

Q8) \(x + 6\over 3\) ÷ \({ x + 2} \over 7 \) = [ \(7( x + 6) \over 3 ( x + 2)\) ]

Q8) \(x + 4\over 3\) ÷ \( x + 4\over x + 6\) = [ \(x + 6\over 3\) ]

Q9) \(x + 4\over 9\) ÷ \(5 \over {x + 2}\) = [ \(x^2 + 6 x + 8\over 45\) ]

Q9) \(x + 5\over 1\) ÷ \({ x + 8} \over 1 \) = [ \(1( x + 5) \over( x + 8)\) ]

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

Q10) \(x + 5\over 5\) x \(x + 6\over 7\) = [ \(x^2 + 11x + 30\over 35\) ]

Q10) \(x + 7\over 8\) ÷ \({ x + 6} \over 3 \) = [ \(3( x + 7) \over 8 ( x + 6)\) ]

Q10) \(x + 7\over 7\) x \(x + 1\over x + 7\) = [ \(x + 1\over 7\) ]