em chưa học hóa thưa chị

nè ko có cưng ở đây nha 2k6 còn lắm chuyện

Ta có: \(\frac{4x^5y^2+2x^4y-6x^3y^2+3xy^4-y^5}{2x^2+xy-y^2}\)

\(=\frac{y\left(4x^5y+2x^4-6x^3y+3xy^3-y^4\right)}{\left(2x-y\right)\left(x+y\right)}\)

Tính ra nữa được k ạ @@

bạn đợi mình tí nhé, khoảng 15 phút sau mình gửi câu trả lời :>>

\(\text{Câu 1 thì bạn tự làm chịu khó tí:}y^2-4x^2=\left(y+2x\right)\left(y-2x\right)\)

\(\text{Bài 2:}ab\left(b-a\right)+bc\left(c-b\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left(b-c\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left[\left(b-a\right)+\left(a-c\right)\right]+ac\left(a-c\right)\)

\(=\left(ab-bc\right)\left(b-a\right)+\left(ac-bc\right)\left(a-c\right)\)

\(=b\left(a-c\right)\left(b-a\right)+c\left(a-b\right)\left(a-c\right)=\left(b-c\right)\left(b-a\right)\left(a-c\right)\)

\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x^3+1\right)\left(x+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)

\(x^4-x^3-x+1=\left(x^4-x^3\right)-\left(x-1\right)=\left(x^3-1\right)\left(x-1\right)=\left(x-1\right)^2\left(x^2+x+1\right)\)

\(x^2y+xy^2-\left(x+y\right)=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)

\(ax^2+ay-bx^2-by=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)

1) x² + x - y² + y = (x² - y²) + (x + y) = (x - y)(x + y) + (x + y) = (x - y + 1)(x + y)

2) 4x² - 9y² + 4x - 6y = (4x² - 9y²) + (4x - 6y) = (2x - 3y)(2x + 3y) + 2(2x - 3y) = (2x - 3y)(2x + 3y + 2)

3) x² + x + y² + y + 2xy = (x² + 2xy + y²) + (x + y) = (x + y)² + (x + y) = (x + y)(x + y + 1)

4) -x² + 5x + 2xy - 5y - y² = -(x² - 2xy + y²) + (5x - 5y) = -(x - y)² + 5(x - y) = (x - y)(y - x + 5)

5) x² - y² + 2x + 1  = (x² + 2x + 1) - y² = (x + 1)² - y² = (x + 1 + y)(x - y + 1)

6) x² - 1 - y² + 2y = x² - (y² - 2y + 1) = x² - (y - 1)² = (x - y + 1)(x + y - 1)

7) x² + 2xz - y² + 2uy + z² - u² =(x² + 2xz + z²) - (y² - 2uy + u²) = (x + z)² - (y - u)² = (x + z - y + u)(x + z + y - u)

8) x³ + 3x²y + x + 3xy² + y + y³ = (x³ + 3x²y + 3xy² + y³) + (x + y) = (x + y)³ + (x + y) = (x + y)(x² + 2xy + y² + 1)

9) x³ + y(1 - 3x²) + x(3y² - 1) - y³ = x³ + y - 3x²y + 3xy² - x - y³ = (x³ - 3x²y + 3xy² - y³) - (x - y) = (x - y)³ - (x - y) = (x - y)(x² - 2xy+y²-1)

Ta có: \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{b^2-a^2}{\left(a+b\right)\left(a+c\right)}+\frac{a^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{b-a}{a+c}+\frac{a-c}{a+b}\left(1\right)\)

Tương tự ta có:

\(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c-b}{a+b}+\frac{b-a}{b+c}\left(2\right)\)

\(\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\left(3\right)\)

(1)(2)(3) => ĐPCM

1)2xy+3z+6y+xz 

= x(2y + z) + 3(z + 2y)

= (x + 3)(2y + z)

2)x^4-9x^3+x^2-9x 

= x^2(x^2 + 1) - 9x(x^2 + 1)

= (x^2 + 1)(x^2 - 9x)

= x(x - 9)(x^2 + 1)

3)x^2-xy+x-y 

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

= (x + 1)(x - y)

4)xz+yz-5(x+y)

= z(x + y) - 5(x + y)

= (z - 5)(x + y)

5)3x^2-3xy-5x+5y 

= 3x(x - y) - 5(x - y)

= (3x - 5)(x - y)

6)x^2+4x-y^2+4y 

= (x - y)(x + y) + 4(x + y)

= (x - y + 4)(x + y)