Question

Разложить на множители 1) (x-y)^2+x-y,. 2)x+y-x^2-xy, 3)4ab^2+5ab+a

Answer 1

1) (x-y)²+x-y=(x-y)²+(x-y)=(x-y)(x-y+1)
2) x+y-x²-xy=(x+y)-(x²+xy)=(x+y)-x(x+y)=(x+y)(1-x)
3) 4ab²+5ab+a=4ab²+4ab+ab+a=(4ab²+4ab)+(ab+a)=4ab(b+1)+a(b+1)=(b+1)(4ab+a)=(b+1)*a(4b+1)=a(b+1)(4b+1)

Answer 2

Разложение на множители:

1) (x-y)^2 + x-y

To factorize this expression, we can observe that it has a common binomial factor of (x-y). By factoring out (x-y), we get:

(x-y)((x-y) + 1)

So, the expression (x-y)^2 + x-y can be factored as (x-y)((x-y) + 1).

2) x+y-x^2-xy

To factorize this expression, we can rearrange the terms and factor by grouping:

x+y-x^2-xy = (x-y) + (y-x^2)

Now, let's factor out the common factors from each group:

(x-y) + (y-x^2) = -(x-y) + x(1-x)

Notice that we can rewrite -(x-y) as (y-x), so the expression becomes:

(y-x) + x(1-x)

Now, let's factor out the common factor of (1-x):

(y-x) + x(1-x) = (1-x)(y-x) + x(1-x)

So, the expression x+y-x^2-xy can be factored as (1-x)(y-x) + x(1-x).

3) 4ab^2 + 5ab + a

To factorize this expression, we can observe that it has a common factor of a. By factoring out a, we get:

a(4b^2 + 5b + 1)

Now, let's factorize the quadratic expression 4b^2 + 5b + 1:

4b^2 + 5b + 1 can be factored as (4b + 1)(b + 1)

So, the expression 4ab^2 + 5ab + a can be factored as a(4b + 1)(b + 1).

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