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Answer :
(i)Given: \(5(2x + 1)(3x + 5) ÷ (2x + 1)\)
\(=\frac{5(2x + 1)(3x + 5)}{2x + 1}=\frac{5\times (2x+3)\times(3x + 5)}{(2x + 1)}\)
\(=\frac{5\times (3x+5)}1=5(3x+5)\)
(ii)Given: \(26xy (x + 5) (y – 4) ÷ 13x (y – 4)\)
\(=\frac{26xy (x + 5) (y – 4)}{13x (y – 4)}=\frac{13\times 2\times x \times y \times (x+5) \times (y-4)}{13\times x\times (y-4)}\)
\(=\frac{2\times y \times (x+5))}1=2y(x+5)\)
(iii)Given: \(52pqr (p + q) (q + r) (r + p) ÷104pq (q + r) (r + p)\)
\(=\frac{52pqr (p + q) (q + r) (r + p)}{104pq (q + r) (r + p)}=\frac{52\times p \times q \times r \times (p+q) \times (q+r) \times (r+p)}{52\times 2 \times p \times q \times (q+r) \times (r+p)}\)
\(=\frac{1}2r(p+q)\)
(iv)Given: \(20 (y + 4) (y^2 + 5y + 3) ÷ 5 (y + 4)\)
\(=\frac{20 (y + 4) (y^2 + 5y + 3) }{5 (y + 4)}=\frac{5\times 4 \times (y+4) \times (y^2+5y+3)}{13\times x\times (y-4)}\)
\(=\frac{4\times (y^2 + 5y + 3)}1=4(y^2 + 5y + 3)\)
(v)Given: \(x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)\)
\(=\frac{x (x + 1) (x + 2) (x + 3) }{x (x + 1)}=\frac{z \times (x+1) \times (x+2) \times (x+3)}{x \times (x+1)}\)
\(=\frac{(x+2)(x+3)}1=(x+2)(x+3)\)