Question

СРОЧНО ПОМОГИТЕ 1. 1)(x+4)^2 2)(a-2b)^2 3)(3y+5)(3y-5) 2. 1)(c-2)(c+3)-(c-1)^2 2)3(a+c)^2-6ac 3. 1)16a^2-9 2)3x^3-75x 3)2x^2+4xy+2y^2 4. (6x-x^2)^2+x^2(x-1)(x+1)+6x(3+2^2)

Answer 1

Пошаговое объяснение....

Answer 2

Sure, I'd be happy to help you simplify these expressions. Let's go through them step by step:

a) $(x+4)^2$ To simplify this expression, you can apply the formula for squaring a binomial: $(a + b)^2 = a^2 + 2ab + b^2$. In this case, $a = x$ and $b = 4$: $(x + 4)^2 = x^2 + 2x \cdot 4 + 4^2 = x^2 + 8x + 16$.

b) $(a-2b)^2$ Using the same formula: $(a - 2b)^2 = a^2 - 2ab + (2b)^2 = a^2 - 4ab + 4b^2$.

c) $(3y+5)(3y-5)$ This expression is a difference of squares, and you can use the formula $a^2 - b^2 = (a + b)(a - b)$: $(3y + 5)(3y - 5) = (3y)^2 - 5^2 = 9y^2 - 25$.

a) $(c-2)(c+3)-(c-1)^2$ Start by expanding the terms inside the brackets and then subtracting the squared term: $(c - 2)(c + 3) - (c - 1)^2 = c^2 + 3c - 2c - 6 - (c^2 - 2c + 1) = c^2 + 3c - 2c - 6 - c^2 + 2c - 1 = 3 - 6 - 1 = -4$.

b) $3(a+c)^2-6ac$ To simplify this expression, first expand the square: $3(a + c)^2 = 3(a^2 + 2ac + c^2) = 3a^2 + 6ac + 3c^2$. Now, subtract $6ac$: $3(a+c)^2 - 6ac = 3a^2 + 6ac + 3c^2 - 6ac = 3a^2 + 3c^2$.

a) $16a^2 - 9$ This expression cannot be further simplified. It is a difference of squares, but $16a^2$ cannot be factored further, and 9 is a perfect square.

b) $3x^3 - 75x$ Here, you can factor out $3x$ from both terms: $3x^3 - 75x = 3x(x^2 - 25)$. Now, you can further factor the difference of squares: $3x(x^2 - 25) = 3x(x + 5)(x - 5)$.

c) $2x^2 + 4xy + 2y^2$ This expression cannot be factored further. It is a quadratic expression with two terms, and no common factors can be extracted.

$(6x - x^2)^2 + x^2(x - 1)(x + 1) + 6x(3 + 2^2)$