БЕРИ:

Решение:

(5x+4y)² = (5x)²+2×5x×4y+(4y)² = 25x²+40xy+16y²

(8a-5b)² = (8a)²-2×8a×5b+(5b)² = 64a²-80ab+25b²

9x²+42xy+49y² = (3x)²+2×3x×7y+(7y)² = (3x+7y)²

64x²-48xy+9y² = (8x)²-2×8x×3y+(3y)² = (8x-3y)²

121x²-0,16y⁴ = (11x)²-(0,4y²)² = (11x-0,4y²)(11x+0,4y²)

(2n-3m)(3m+2n) = (2n-3m)(2n+3m) = (2n)²-(3m)² = 4n²-9m²

125x³+216y³ = (5x)³+(6y)³ = (5x+6y)((5x)²-5x×6y+(6y)²) = (5x+6y)(25x²-30xy+36y²)

27a³-64b³ = (3a)³-(4b)³ = (3a-4b)((3a)²+3a×4b+(4b)²) = (3a-4b)(9a²+12ab+16b²)