Question

Превратите в стандартный вид многочлены а)(х-3)² б) (х-2)(х+2) в) (2а+ 5в)² г) (3х-у)(у+3х)

Answer 1

А. х²-6х+9
Б. х²-4
В. 4а²+20аб+25б²
Г. 9х²-у²

Answer 2

Ответы 100% правильные!!!!

Answer 3

a) (x-3)²

To expand the expression (x-3)², we can use the formula for expanding a binomial squared, which is (a-b)² = a² - 2ab + b². In this case, a = x and b = 3. Plugging these values into the formula, we get:

(x-3)² = x² - 2(x)(3) + 3² = x² - 6x + 9

So, the expanded form of (x-3)² is x² - 6x + 9.

б) (x-2)(x+2)

To expand the expression (x-2)(x+2), we can use the distributive property. We multiply each term in the first expression (x-2) by each term in the second expression (x+2). Applying this rule, we get:

(x-2)(x+2) = x(x) + x(2) - 2(x) - 2(2) = x² + 2x - 2x - 4 = x² - 4

So, the expanded form of (x-2)(x+2) is x² - 4.

в) (2a + 5b)²

To expand the expression (2a + 5b)², we can use the formula for expanding a binomial squared, which is (a+b)² = a² + 2ab + b². In this case, a = 2a and b = 5b. Plugging these values into the formula, we get:

(2a + 5b)² = (2a)² + 2(2a)(5b) + (5b)² = 4a² + 20ab + 25b²

So, the expanded form of (2a + 5b)² is 4a² + 20ab + 25b².

г) (3x - y)(y + 3x)

To expand the expression (3x - y)(y + 3x), we can again use the distributive property. We multiply each term in the first expression (3x - y) by each term in the second expression (y + 3x). Applying this rule, we get:

(3x - y)(y + 3x) = (3x)(y) + (3x)(3x) - (y)(y) - y(3x) = 3xy + 9x² - y² - 3xy = 9x² - y²

So, the expanded form of (3x - y)(y + 3x) is 9x² - y².

I hope this helps! Let me know if you have any further questions.