Question

Представить в виде многочлена (раскрыть скобки) 1). (x - 2)(x + 2) 2). (3x + y)(3x - y) 3). (a² - 3)(a² + 3) 4). (5 - y)² 5). (x + 7)2 + (x + 2)(x - 2) 6). (a + 1)(a - 1) – (a + 4)2 7). (x - 5)2 - (x - 7)(x + 7) 8). (c-y)(c + y) (c² + y²)​

Answer 1

Ответ:

1). x2+2x-2x-4=x2-4 2).9x2-3xy+3xy-y2=9x2-y2 3). a4 +3a2-3a2-9=a4-9 4). 25-10y+y2 5). 2x+14+x2-2x+2x-4=4x+10

то что x2 и т.д это x в квадрате

Answer 2

1) (x - 2)(x + 2)

To expand this expression, we can use the FOIL method, which stands for First, Outer, Inner, Last.

First, we multiply the first terms of each binomial: x * x = x^2

Outer, we multiply the outer terms of each binomial: x * 2 = 2x

Inner, we multiply the inner terms of each binomial: -2 * x = -2x

Last, we multiply the last terms of each binomial: -2 * 2 = -4

Now, we can combine the like terms: x^2 + 2x - 2x - 4

The middle terms, 2x and -2x, cancel each other out, so we are left with: x^2 - 4

Therefore, the expanded form of (x - 2)(x + 2) is x^2 - 4.

2) (3x + y)(3x - y)

To expand this expression, we can once again use the FOIL method:

First, we multiply the first terms of each binomial: 3x * 3x = 9x^2

Outer, we multiply the outer terms of each binomial: 3x * -y = -3xy

Inner, we multiply the inner terms of each binomial: y * 3x = 3xy

Last, we multiply the last terms of each binomial: y * -y = -y^2

Now, we can combine the like terms: 9x^2 - 3xy + 3xy - y^2

The middle terms, -3xy and 3xy, cancel each other out, so we are left with: 9x^2 - y^2

Therefore, the expanded form of (3x + y)(3x - y) is 9x^2 - y^2.

3) (a² - 3)(a² + 3)

To expand this expression, we can once again use the FOIL method:

First, we multiply the first terms of each binomial: a² * a² = a^4

Outer, we multiply the outer terms of each binomial: a² * 3 = 3a²

Inner, we multiply the inner terms of each binomial: -3 * a² = -3a²

Last, we multiply the last terms of each binomial: -3 * 3 = -9

Now, we can combine the like terms: a^4 + 3a² - 3a² - 9

The middle terms, 3a² and -3a², cancel each other out, so we are left with: a^4 - 9

Therefore, the expanded form of (a² - 3)(a² + 3) is a^4 - 9.

4) (5 - y)²

To expand this expression, we can again use the FOIL method:

First, we multiply the first terms of each binomial: 5 * 5 = 25

Outer, we multiply the outer terms of each binomial: 5 * -y = -5y

Inner, we multiply the inner terms of each binomial: -y * 5 = -5y

Last, we multiply the last terms of each binomial: -y * -y = y²

Now, we can combine the like terms: 25 - 5y - 5y + y²

The middle terms, -5y and -5y, combine to give us -10y, so we are left with: 25 - 10y + y²

Therefore, the expanded form of (5 - y)² is 25 - 10y + y².

5) (x + 7)² + (x + 2)(x - 2)

To expand this expression, we can first expand the square term using the FOIL method:

(x + 7)² = (x + 7)(x + 7)

First, we multiply the first terms of each binomial: x * x = x²

Outer, we multiply the outer terms of each binomial: x * 7 = 7x

Inner, we multiply the inner terms of each binomial: 7 * x = 7x

Last, we multiply the last terms of each binomial: 7 * 7 = 49

Now, we can combine the like terms: x² + 7x + 7x + 49

The middle terms, 7x and 7x, combine to give us 14x, so we are left with: x² + 14x + 49

Next, we expand the second term (x + 2)(x - 2) using the FOIL method:

First, we multiply the first terms of each binomial: x * x = x²

Outer, we multiply the outer terms of each binomial: x * -2 = -2x

Inner, we multiply the inner terms of each binomial: 2 * x = 2x

Last, we multiply the last terms of each binomial: 2 * -2 = -4

Now, we can combine the like terms: x² - 2x + 2x - 4

The middle terms, -2x and 2x, cancel each other out, so we are left with: x² - 4

Finally, we add the two expanded terms together: (x + 7)² + (x + 2)(x - 2) = x² + 14x + 49 + x² - 4

Combining like terms, we get: 2x² + 14x + 45

Therefore, the expanded form of (x + 7)² + (x + 2)(x - 2) is 2x² + 14x + 45.

6) (a + 1)(a - 1) - (a + 4)²

To expand this expression, we can first expand the first term (a + 1)(a - 1) using the FOIL method:

First, we multiply the first terms of each binomial: a * a = a²

Outer, we multiply the outer terms of each binomial: a * -1 = -a

Inner, we multiply the inner terms of each binomial: 1 * a = a

Last, we multiply the last terms of each binomial: 1 * -1 = -1

Now, we can combine the like terms: a² - a + a - 1

The middle terms, -a and a, cancel each other out, so we are left with: a² - 1

Next, we expand the second term (a + 4)²:

(a + 4)² = (a + 4)(a + 4)

Using the FOIL method, we multiply the binomials:

First, we multiply the first terms of each binomial: a * a = a²

Outer, we multiply the outer terms of each binomial: a * 4 = 4a

Inner, we multiply the inner terms of each binomial: 4 * a = 4a

Last, we multiply the last terms of each binomial: 4 * 4 = 16

Now, we can combine the like terms: a² + 4a + 4a + 16

The middle terms, 4a and 4a, combine to give us 8a, so we are left with: a² + 8a + 16

Finally, we subtract the expanded second term from the expanded first term: (a + 1)(a - 1) - (a + 4)² = (a² - 1) - (a² + 8a + 16)

Distributing the negative sign to each term in the second parentheses, we get: a² - 1 - a² - 8a - 16

Combining like terms, we have: -8a - 17

Therefore, the expanded form of (a + 1)(a - 1) - (a + 4)² is -8a - 17.

7) (x - 5)² - (x - 7)(x + 7)

To expand this expression, we can first expand the square term using the FOIL method:

(x - 5)² = (x - 5)(x - 5)

First, we multiply the first terms of each binomial: x * x = x²

Outer, we multiply the outer terms of each binomial: x * -5 = -5x

Inner, we multiply the inner terms of each binomial: -5 * x = -5x

Last, we multiply the last terms of each binomial: -5 * -5 = 25

Now, we can combine the like terms: x² - 5x - 5x + 25

The middle terms, -5x and -5x, combine to give us -10x, so we are left with: x² - 10x + 25

Next, we expand the second term (x - 7)(x + 7) using the FOIL method:

First, we multiply the first terms of each binomial: x * x = x²

Outer, we multiply the outer terms of each binomial: x * 7 = 7x

Inner, we multiply the inner terms of each binomial: -7 * x = -7x

Last, we multiply the last terms of each binomial: