Q1) \(x + 10\over 7\) ÷ \(7 \over {x + 4}\) = [ \(x^2 + 14 x + 40\over 49\) ]

Q1) \(x + 4\over 1\) x \(1 \over{ x + 8}\) = [ \(1( x + 4) \over( x + 8)\) ]

Q1) \(x + 1\over 3\) x \(x + 3\over x + 1\) = [ \(x + 3\over 3\) ]

Q2) \(x + 2\over 4\) x \(x + 1\over 3\) = [ \(x^2 + 3 x + 2\over 12\) ]

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

Q2) \(x + 5\over 9\) ÷ \( x + 5\over x + 8\) = [ \(x + 8\over 9\) ]

Q3) \(x + 7\over 7\) x \(x + 5\over 3\) = [ \(x^2 + 12 x + 35\over 21\) ]

Q3) \(x + 5\over 9\) x \(4 \over{ x + 4}\) = [ \(4( x + 5) \over 9 ( x + 4)\) ]

Q3) \(x + 6\over 6\) ÷ \( x + 6\over x + 2\) = [ \(x + 2\over 6\) ]

Q4) \(x + 8\over 5\) x \(x + 3\over 5\) = [ \(x^2 + 11x + 24\over 25\) ]

Q4) \(x + 10\over 9\) ÷ \({ x + 9} \over 5 \) = [ \(5( x + 10) \over 9 ( x + 9)\) ]

Q4) \(x + 10\over 8\) ÷ \( x + 10\over x + 10\) = [ \(x + 10\over 8\) ]

Q5) \(x + 3\over 9\) x \(x + 8\over 3\) = [ \(x^2 + 11x + 24\over 27\) ]

Q5) \(x + 8\over 8\) x \(5 \over{ x + 6}\) = [ \(5( x + 8) \over 8 ( x + 6)\) ]

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

Q6) \(x + 1\over 4\) x \(x + 1\over 3\) = [ \(x^2 + 2 x + 1\over 12\) ]

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

Q6) \(x + 6\over 2\) x \(x + 7\over x + 6\) = [ \(x + 7\over 2\) ]

Q7) \(x + 7\over 5\) ÷ \(5 \over {x + 3}\) = [ \(x^2 + 10 x + 21\over 25\) ]

Q7) \(x + 9\over 10\) ÷ \({ x + 4} \over 9 \) = [ \(9( x + 9) \over 10 ( x + 4)\) ]

Q7) \(x + 6\over 10\) ÷ \( x + 6\over x + 4\) = [ \(x + 4\over 10\) ]

Q8) \(x + 9\over 3\) x \(x + 7\over 8\) = [ \(x^2 + 16 x + 63\over 24\) ]

Q8) \(x + 1\over 1\) ÷ \({ x + 8} \over 1 \) = [ \(1( x + 1) \over( x + 8)\) ]

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

Q9) \(x + 10\over 9\) x \(x + 8\over 4\) = [ \(x^2 + 18 x + 80\over 36\) ]

Q9) \(x + 4\over 4\) x \(5 \over{ x + 6}\) = [ \(5( x + 4) \over 4 ( x + 6)\) ]

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

Q10) \(x + 4\over 7\) ÷ \(10 \over {x + 2}\) = [ \(x^2 + 6 x + 8\over 70\) ]

Q10) \(x + 1\over 2\) ÷ \({ x + 2} \over 7 \) = [ \(7( x + 1) \over 2 ( x + 2)\) ]

Q10) \(x + 4\over 6\) x \(x + 9\over x + 4\) = [ \(x + 9\over 6\) ]