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Ответ:
Решение:
Применяем формулу разности квадратов \( (a-b)(a+b) = a^2-b^2 \).
- \( (x - y)(x + y) = x^2 - y^2 \)
- \( (a-b)(a+b) = a^2 - b^2 \)
- \( (n - m)(n + m) = n^2 - m^2 \)
- \( (c - k)(c + k) = c^2 - k^2 \)
- \( (b-3)(b+3) = b^2 - 3^2 = b^2 - 9 \)
- \( (h-7)(h+7) = h^2 - 7^2 = h^2 - 49 \)
- \( (s+2)(s-2) = s^2 - 2^2 = s^2 - 4 \)
- \( (9-n)(9+n) = 9^2 - n^2 = 81 - n^2 \)
- \( (x-1)(x+1) = x^2 - 1^2 = x^2 - 1 \)
- \( (2-z)(2+z) = 2^2 - z^2 = 4 - z^2 \)
- \( (1+p)(1-p) = 1^2 - p^2 = 1 - p^2 \)
- \( (2x+5)(2x-5) = (2x)^2 - 5^2 = 4x^2 - 25 \)
- \( (d+3c)(d-3c) = d^2 - (3c)^2 = d^2 - 9c^2 \)
- \( (1-4y)(1+4y) = 1^2 - (4y)^2 = 1 - 16y^2 \)
- \( (7a+2b)(7a-2b) = (7a)^2 - (2b)^2 = 49a^2 - 4b^2 \)
- \( (10x-y)(10x+y) = (10x)^2 - y^2 = 100x^2 - y^2 \)
- \( (0,2n+y^2)(0,2n-y^2) = (0,2n)^2 - (y^2)^2 = 0,04n^2 - y^4 \)
- \( (5a^2-0,6b)(5a^2+0,6b) = (5a^2)^2 - (0,6b)^2 = 25a^4 - 0,36b^2 \)
- \( (5r-3)(3+5r) = (5r-3)(5r+3) = (5r)^2 - 3^2 = 25r^2 - 9 \)
- \( (8p-3)(3+8p) = (8p-3)(8p+3) = (8p)^2 - 3^2 = 64p^2 - 9 \)
- \( (k-2m)(2m+k) = (k-2m)(k+2m) = k^2 - (2m)^2 = k^2 - 4m^2 \)
- \( (s+9z)(9z-s) = (9z+s)(9z-s) = (9z)^2 - s^2 = 81z^2 - s^2 \)
- \( (6h+5)(5-6h) = (5+6h)(5-6h) = 5^2 - (6h)^2 = 25 - 36h^2 \)
- \( (q-7)(7+q) = (q-7)(q+7) = q^2 - 7^2 = q^2 - 49 \)
- \( (b+2a)(b-2a) = b^2 - (2a)^2 = b^2 - 4a^2 \)
- \( (cd+1)(1-cd) = (1+cd)(1-cd) = 1^2 - (cd)^2 = 1 - c^2d^2 \)
- \( (2y-5z)(2y+5z) = (2y)^2 - (5z)^2 = 4y^2 - 25z^2 \)
- \( (1,2x+hpc)(hpc-1,2x) = (hpc+1,2x)(hpc-1,2x) = (hpc)^2 - (1,2x)^2 = hpc^2 - 1,44x^2 \)
- \( (10b-0,1d)(0,1d+10b) = (10b-0,1d)(10b+0,1d) = (10b)^2 - (0,1d)^2 = 100b^2 - 0,01d^2 \)
- \( (1,2n+4y^2)(4y^2-1,2n) = (4y^2+1,2n)(4y^2-1,2n) = (4y^2)^2 - (1,2n)^2 = 16y^4 - 1,44n^2 \)
- \( (5q^5-0,5b^3)(5q^5+0,5b^3) = (5q^5)^2 - (0,5b^3)^2 = 25q^{10} - 0,25b^6 \)
- \( (2x^2y+30n^2)(30n^2-2x^3y) \) = \( (30n^2+2x^2y)(30n^2-2x^2y) = (30n^2)^2 - (2x^2y)^2 = 900n^4 - 4x^4y^2 \)
