Question

A) 16x²-9= б) 3x³-75x= в) 2x²+4xy+2y²=

Answer 1

Формулы:

1. Разность квадратов: a²–b²= (a–b)(a+b)

2. Квадрат суммы: (a+b)²= a²+2ab+b².

а) 16x²– 9= (4х–3)(4х+3).

б) 3x³– 75x= 3х(х²–25)= 3х(х–5)(х+5).

в) 2x²+4xy+2y²= 2(х²+2ху+у²)= 2(х+у)².

Answer 2

а)

$\displaystyle \tt 16x^2-9=4^2x^2-3^2=(4x)^2-3^2=\bold{(4x-3)(4x+3)}$

б)

$\displaystyle \tt 3x^3-75x=3x(x^2-25)=\bold{3x(x-5)(x+5)}$

в)

$\displaystyle \tt 2x^2+4xy+2y^2=2(x^2+2xy+y^2)=2(x+y)^2=\bold{2(x+y)(x+y)}$

Answer 3

To solve these equations, we need to find the values of 'x' that satisfy each equation. Let's solve them one by one:

A) 16x² - 9 = 0

Step 1: Move the constant term to the other side. 16x² = 9

Step 2: Divide both sides by 16. x² = 9/16

Step 3: Take the square root of both sides. x = ±√(9/16)

x = ±(3/4)

So the solutions for A) are x = 3/4 and x = -3/4.

B) 3x³ - 75x = 0

Step 1: Factor out the common term '3x'. 3x(x² - 25) = 0

Step 2: Factor the difference of squares (x² - 25). 3x(x + 5)(x - 5) = 0

Step 3: Set each factor to zero and solve for 'x'.

1. 3x = 0 --> x = 0
2. x + 5 = 0 --> x = -5
3. x - 5 = 0 --> x = 5

So the solutions for B) are x = 0, x = -5, and x = 5.

C) 2x² + 4xy + 2y² = 0

This is a quadratic equation in two variables 'x' and 'y'. It represents a conic section. However, to find specific solutions, we need additional information or constraints.

If you have specific values for 'x' and 'y' or any additional equation or constraints related to 'x' and 'y', please provide them, and I'll be happy to help you find the solutions. Otherwise, without further information, we cannot find specific solutions for C).