\(B(y) - A(y) = 2{y^3} - 9{y^2} + 4y\)

\(\begin{array}{l}A(y) =  - 5{y^4} - 4{y^2} + 2y + 7\\ \Rightarrow B(y) = 2{y^3} - 9{y^2} + 4y - 5{y^4} - 4{y^2} + 2y + 7\\ =  - 5{y^4} + 2{y^3} - 13{y^2} + 6y + 7\end{array}\)

\(\left(1+2\right),y^2-13y+12=y^2-12y-y-12=y\left(y-12\right)+\left(y-12\right)=\left(y+1\right)\left(y-12\right)\)

\(3,x^2-x-30=x^2-6x+5x-30=x\left(x-6\right)+5\left(x-6\right)=\left(x+5\right)\left(x-6\right)\)

\(4,y^2+y-42=y^2-6y+7y-42=y\left(y-6\right)+7\left(y-6\right)=\left(y+7\right)\left(y-6\right)\)

\(5,x^2+3x-10=x^2-2x+5x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)

\(6,x^2-8x+15=x^2-5x-3x+15=x\left(x-5\right)-3\left(x-5\right)=\left(x-3\right)\left(x-5\right)\)

Đơn giản

\(C=\dfrac{2}{3}y^4-\dfrac{3}{4}y^2-\dfrac{9}{2}y-\dfrac{1}{3}y^3+y-\dfrac{2}{5}=-\dfrac{2}{5}-\dfrac{7}{2}y-\dfrac{3}{4}y^2-\dfrac{1}{3}y^3+\dfrac{2}{3}y^4\)\(D=\dfrac{1}{5}y^2-y-3y^4+2=2-y+\dfrac{1}{5}y^2-3y^4\)

Bây giờ tìm hiệu của C - D

Ta có hiệu của C - D bằng: ( xin lỗi vì không biết cách đặt tính )

\(C=-\dfrac{2}{5}-\dfrac{7}{2}y-\dfrac{3}{4}y^2-\dfrac{1}{3}y^3+\dfrac{2}{3}y^4-D=2-y+\dfrac{1}{5}y^2-3y^4\)

C - D = \(-\dfrac{12}{5}-\dfrac{5}{2}y-\dfrac{19}{20}y^2-\dfrac{1}{3}y^3+\dfrac{11}{3}y^4\)

Easy

??                       

Đề lỗi rồi kìa, bạn viết lại đi tridung

`a)`

`A=-4x^5y^3+6x^4y^3-3x^2y^3z^2+4x^5y^3-x^4y^3+3x^2y^3z^2-2y^4+22`

`A=(-4x^5y^3+4x^5y^3)+(6x^4y^3-x^4y^3)-(3x^2y^3z^2-3x^2y^3z^2)-2y^4+22`

`A=5x^4y^3-2y^4+22`

        `->` Bậc: `7`

`b)B-5y^4=A`

`=>B=A+5y^4`

`=>B=5x^4y^3-2y^4+22+5y^4`

`=>B=5x^4y^3+3y^4+22`

a) Thay x = -1, y = 1 vào đa thức A ta được:

\(\begin{array}{l}A = 4.{\left( { - 1} \right)^6} - 2.{\left( { - 1} \right)^2}{.1^3} - 5.\left( { - 1} \right).1 + 2\\A = 4 - 2 + 5 + 2 = 9\end{array}\)

Vậy A =9 tại x = -1; y = 1

Thay x = -1, y = 1 vào đa thức B ta được:

\(\begin{array}{l}B = 3.{\left( { - 1} \right)^2}{.1^3} + 5.\left( { - 1} \right).1 - 7\\B = 3 - 5 - 7 =  - 9\end{array}\)

Vậy B = -9 tại x = -1; y = 1

b) Ta có:

\(\begin{array}{l}A + B = \left( {4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2} \right) + \left( {3{{\rm{x}}^2}{y^3} + 5{\rm{x}}y - 7} \right)\\ = 4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2 + 3{{\rm{x}}^2}{y^3} + 5{\rm{x}}y - 7\\ = 4{{\rm{x}}^6} + \left( { - 2{{\rm{x}}^2}{y^3} + 3{{\rm{x}}^2}{y^3}} \right) + \left( { - 5{\rm{x}}y + 5{\rm{x}}y} \right) + 2 - 7\\ = 4{{\rm{x}}^6} + {x^2}{y^3} - 5\end{array}\)

\(\begin{array}{l}A - B = \left( {4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2} \right) - \left( {3{{\rm{x}}^2}{y^3} + 5{\rm{x}}y - 7} \right)\\ = 4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2 - 3{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 7\\ = 4{{\rm{x}}^6} + \left( { - 2{{\rm{x}}^2}{y^3} - 3{{\rm{x}}^2}{y^3}} \right) + \left( { - 5{\rm{x}}y - 5{\rm{x}}y} \right) + 2 + 7\\ = 4{{\rm{x}}^6} - 5{x^2}{y^3} - 10{\rm{x}}y + 9\end{array}\)

Ta có:

\(\begin{array}{l}(12{{\rm{x}}^3}{y^3} - 6{{\rm{x}}^4}{y^3} + 21{{\rm{x}}^3}{y^4}):(3{{\rm{x}}^3}{y^3})\\ = (12{{\rm{x}}^3}{y^3}):\left( {3{{\rm{x}}^3}{y^3}} \right) + \left( { - 6{{\rm{x}}^4}{y^3}} \right):\left( {3{{\rm{x}}^3}{y^3}} \right) + \left( {21{{\rm{x}}^3}{y^4}} \right):\left( {3{{\rm{x}}^3}{y^3}} \right)\\ = 4 - 2{\rm{x}} + 7y\end{array}\)

Thương của phép chia đa thức\(12{{\rm{x}}^3}{y^3} - 6{{\rm{x}}^4}{y^3} + 21{{\rm{x}}^3}{y^4}\) cho đơn thức \(3{{\rm{x}}^3}{y^3}\) là 4 – 2x +7y

1: \(3x\left(x-1\right)+7x^2\left(x-1\right)\)

\(=\left(x-1\right)\left(7x^2+3x\right)\)

=x(7x+3)(x-1)

2: \(9\left(x+5\right)^2-\left(x-7\right)^2\)

\(=\left(3x+15\right)^2-\left(x-7\right)^2\)

=(3x+15+x-7)(3x+15-x+7)

=(4x+8)(2x+22)

\(=4\left(x+2\right)\cdot2\cdot\left(x+11\right)=8\left(x+2\right)\left(x+11\right)\)

3: \(x^2-2xy+y^2-xz+yz\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

=(x-y)(x-y-z)

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

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

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

5: \(9x^2y^2+15x^2y-21xy^2\)

\(=3xy\cdot3xy+3xy\cdot5x-3xy\cdot7y\)

=3xy(3xy+5x-7y)

6: \(x^3-2x^2y+xy^2-9xy^4\)

\(=x\left(x^2-2xy+y^2-9y^4\right)\)

\(=x\left\lbrack\left(x-y\right)^2-\left(3y^2\right)^2\right\rbrack=x\left(x-y-3y^2\right)\left(x-y+3y^2\right)\)

7: \(x^2+x-12\)

\(=x^2+4x-3x-12\)

=x(x+4)-3(x+4)

=(x+4)(x-3)

8: (x+1)(x+3)(x+5)(x+7)+15

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+15\)

\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)

\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)