`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`

bạn viết có thánh đọc ra á :v

Bạn viết như vậy vẫn nhìn đc nhưng nhìn hơi khó

Thì các bạn vít ra giấy là hỉu nk mong giải giúp mk cái

\(\left(3x+1\right)^2-4\left(x-2\right)^2=9x^2+6x+1-4\left(x^2-4x+4\right)=9x^2+6x+1-4x^2+16x-16=5x^2+22x-15=\)

\(\left(5x-3\right)\left(x+5\right)\)

\(9\left(2x+3\right)^2-4\left(x+1\right)^2=9\left(4x^2+12x+9\right)-4\left(x^2+2x+1\right)=36x^2+108x+81-4x^2-8x-4=32x^2+100x+77\)

\(\left(8x+11\right)\left(4x+7\right)\)

\(8x^3-64=\left(2x\right)^3-4^3=\left(2x-4\right)\left(4x^2+8x+16\right)\)

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

\(3,x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\\ 4,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\\ 5,x^2-2xy+y^2-xz+yz=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y-z\right)\left(x-y\right)\\ 6,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\\ 9,x^3+x^2-xy+xy+y^2+y^3\\ =x^2\left(x+1\right)+y^2\left(x+1\right)=\left(x^2+y^2\right)\left(x+1\right)\\ 10,x^2-6\left(x+3\right)-9\\ =x^2-6x-18-9\\ =x^2-6x-27=\left(x-9\right)\left(x+3\right)\)

10: \(x^2-6\left(x+3\right)-9\)

\(=x^2-6x-18-9\)

\(=x^2-6x-27\)

\(=\left(x-9\right)\left(x+3\right)\)

3)(x-1).(x-y)

a: \(x^2-9-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(1-x^2\right)\)

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

b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)

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

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

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

c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)

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

d: \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

e: \(3x^2-4x-4\)

\(=3x^2-6x+2x-4\)

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

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

g: \(x^4+64y^4\)

\(=x^4+16x^2y^2+64y^4-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

h: \(a^2+b^2+2a-2b-2ab\)

\(=a^2-2ab+b^2+2a-2b\)

\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)

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

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

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

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

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

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

1.

\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+4}{x-3}\)

b.

\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)

\(\Rightarrow x=10\) (thỏa mãn)

2.

\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)

Đề lỗi rồi kìa, bạn viết lại đi tridung