Nhận thấy tử và mẫu phân thức đều có tổng các hệ số bằng 0, Theo Bezout ta có : \(3x^3-7x^2+5x-1⋮x-1,2x^3-x^2-4x+3⋮x-1\)

Thực hiện phép chia ta sẽ có biểu thức

=\(\frac{\left(x-1\right)\left(3x^2-4x+1\right)}{\left(x-1\right)\left(2x^2+x-3\right)}=\frac{3x^2-4x+1}{2x^2+x-3}\).Ta lại thấy tử và mẫu có tổng các hệ số bằng 0, theo Bơ du chúng sẽ chia ht x-1.Thực hiện phép chia rồi rút gọn đc

\(\frac{3x-1}{2x+3}\)

`@` `\text {Ans}`

`\downarrow`

`1.`

\(\left(-4xy\right)\cdot\left(2xy^2-3x^2y\right)\)

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

`=`\(-8\left(x\cdot x\right)\left(y\cdot y^2\right)+12\left(x\cdot x^2\right)\left(y\cdot y\right)\)

`=`\(-8x^2y^3+12x^3y^2\)

`2.`

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

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

`=`\(-15x^4-35x^3+5x^2\)

`3.`

\(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)

`=`\(3x\left(4x+5\right)-2\left(4x+5\right)-12x^2+6x\)

`=`\(12x^2+15x-8x-10-12x^2+6x\)

`=`\(\left(12x^2-12x^2\right)+\left(15x-8x+6x\right)-10\)

`=`\(13x-10\)

`4.`

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

`=`\(2x^2\cdot x^2+2x^2\cdot\left(-7x\right)+2x^2\cdot9\)

`=`\(2x^4-14x^3+18x^2\)

`5.`

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

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

`=`\(3x^3-15x^2+21x-5x^2+25x-35\)

`=`\(3x^3-20x^2+46x-35\)

a: F(x)=3x^3-2x^2+5x-7

G(x)=3x^3-2x^2+5x+7x^2+3=3x^3+5x^2+5x+3

Bậc của F(x),G(x) đều là 3

b: N(x)=G(x)-F(x)

\(=3x^3+5x^2+5x+3-3x^3+2x^2-5x+7=7x^2+10\)

M(x)=2F(x)+G(x)

\(=6x^3-4x^2+10x-14+3x^3+5x^2+5x+3\)

\(=9x^3+x^2+15x-11\)

c: x^2-3x=0

=>x=0 hoặc x=3

\(M\left(0\right)=9\cdot0^3+0^2+15\cdot0-11=-11\)

\(M\left(3\right)=9\cdot3^3+3^2+15\cdot3-11=286\)

d: N(x)=7x^2+10>=10

Dấu = xảy ra khi x=0

\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)

\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)

\(A=7x^3-10x^2+5x-1\)

\(B=-8x^4+7x^2+3\)

a: P(x)=4x-6x^3+5x-7x^2-9x+5

=-6x^3-7x^2+5

Q(x)=9x^3-3x^3+5x^2-4x^2+2x-2x-4

=6x^3+x^2-4

b: P(x)+Q(x)

=-6x^3-7x^2+5+6x^3+x^2-4

=-6x^2+1

P(x)-Q(x)

=-6x^3-7x^2+5-6x^3-x^2+4

=-12x^3-8x^2+9

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

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

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)

1) a)

 \(A\left(x\right)=x^3+5x-7x^2-2x-12+3x^3\\ \text{ }=\left(x^3+3x^3\right)-7x^2+\left(5x-2x\right)-12\\ \text{ }=4x^3-7x^2+3x-12\)

\(B\left(x\right)=-2x^3+2x^2+12+5x^2-9x\\ \text{ }=-2x^3+\left(2x^2+5x^2\right)-9x+12\\ \text{ }=-2x^3+7x^2-9x+12\)

b)

\(A\left(x\right)+B\left(x\right)=\left(4x^3-7x^2+3x-12\right)+\left(-2x^3+7x^2-9x+12\right)\\ \text{ }=4x^3-7x^2+3x-12-2x^3+7x^2-9x+12\\ \text{ }=\left(4x^3-2x^3\right)+\left(7x^2-7x^2\right)-\left(9x-3x\right)+\left(12-12\right)\\ \text{ }=2x^3-6x\)

\(B\left(x\right)-A\left(x\right)=\left(-2x^3+7x^2-9x+12\right)-\left(4x^3-7x^2+3x-12\right)\\ \text{ }=-2x^3+7x^2-9x+12-4x^3+7x^2-3x+12\\ \text{ }=\left(-2x^3-4x^3\right)+\left(7x^2+7x^2\right)-\left(9x+3x\right)+\left(12+12\right)\\ \text{ }=6x^3+14x^2-12x+24\)

\(\left(4x-7\right)\cdot\left(x+5\right)\\ =4x\left(x+5\right)-7\left(x+5\right)\\ =4x\cdot x+4x\cdot5-7\cdot x-7\cdot5\\ =4x^2+20x-7x-35\)

\(-3x^3\left(4x^5+7x^2\right)=-12x^8-21x^5\)