Bài 3L

a: 2021x(x-3)+x-3=0

=>(x-3)(2021x+1)=0

=>\(\left[\begin{array}{l}x-3=0\\ 2021x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac{1}{2021}\end{array}\right.\)

b: \(2x\left(x-2\right)+\left(x+1\right)\left(5-2x\right)=4\)

=>\(2x^2-4x+5x-2x^2+5-2x=4\)

=>-x+5=4

=>-x=-1

=>x=1

Bài 1:

a: \(2x^2+5x-2xy-5y\)

=x(2x+5)-y(2x+5)

=(2x+5)(x-y)

b: \(y\left(x-z\right)+7\left(z-x\right)\)

=y(x-z)-7(x-z)

=(x-z)(y-7)

`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`

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vì x - y - z = 0 nên x = y + z

Xét tổng A + B = xyz - xy² - xz² + y³ + z³

= ( y + z ) . yz - ( y + z ) . y² - ( y + z ) . z² + y³ + z³

= y²z + yz² - y³ - y²z - yz² - z³ + y³ + z³ = 0

Vậy ...

x-y-z=0

=>x=y+z

=>x²=y²+z²+2yz

=>y²+z²=x²-2yz

*A=xyz-xy²-xz²=x.(yz-y²-z²)=x.[yz-(x²-2yz)]=x.(3yz-x²)=3xyz-x³

*B=y³+z³=(y+z)(x²-yz+z²)=x.(x²-2yz-yz)=x³-3xyz=-(3xyz-x³)

Vậy A và B đối nhau