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volslav3
@volslav3
July 2022
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Помогите срочно!!!!!!!!!!!!!!!!!!!!!!!
Разложите на множители (представьте в виде произведения двучленов)
ФСУ. 7 класс
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Verified answer
4 - y² = 2² - y² = (2 - y)(2 + y)
b² - c² = (b - c)(b + c)
4a² - 25 = (2a)² - 5² = (2a - 5)(2a + 5)
25x² - y² = ( 5x)² - y² = (5x - y)(5x + y)
x²y² - 4 = (xy)² - 2² = (xy - 2)(xy + 2)
y⁴ - x² = (y²)² - x² = (y² - x)(y² + x)
25x² - 49y² = (5x)² - (7y)² = (5x - 7y)(5x + 7y)
100 + 25n² = 25(4 + n²)
1,21p² - a⁶ = (1,1p)² - (a³)² = (1,1p - a³)(1,1p + a³)
x² - 1 = (x - 1)(x + 1)
0,25a² - 1 = (0,5a)² - 1² = (0,5a - 1)(0,5a + 1)
100x⁴ - 9y¹⁰ = (10x²)² - (3y⁵)² = (10x² - 3y⁵)(10x² + 3y⁵)
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Answers & Comments
Verified answer
4 - y² = 2² - y² = (2 - y)(2 + y)b² - c² = (b - c)(b + c)
4a² - 25 = (2a)² - 5² = (2a - 5)(2a + 5)
25x² - y² = ( 5x)² - y² = (5x - y)(5x + y)
x²y² - 4 = (xy)² - 2² = (xy - 2)(xy + 2)
y⁴ - x² = (y²)² - x² = (y² - x)(y² + x)
25x² - 49y² = (5x)² - (7y)² = (5x - 7y)(5x + 7y)
100 + 25n² = 25(4 + n²)
1,21p² - a⁶ = (1,1p)² - (a³)² = (1,1p - a³)(1,1p + a³)
x² - 1 = (x - 1)(x + 1)
0,25a² - 1 = (0,5a)² - 1² = (0,5a - 1)(0,5a + 1)
100x⁴ - 9y¹⁰ = (10x²)² - (3y⁵)² = (10x² - 3y⁵)(10x² + 3y⁵)