=x^2-x^2-3x+x+4x^3-1

=4x^3-2x-1

a: \(P\left(x\right)=4x^3+x^2-2x+7\)

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

b: \(M\left(x\right)=4x^3+x^2-2x+7-4x^3-x^2+4x-3=2x+4\)

\(N\left(x\right)=8x^3+2x^2-6x+10\)

c: Đặt M(x)=0

=>2x+4=0

hay x=-2

\(a,Q_{\left(x\right)}=-4x^3+2x-2+2x-x^2-1\\ Q_{\left(x\right)}=-4x^3-x^2+4x-3\\ P_{\left(x\right)}=4x^3-3x+x^2+7+x\\ P_{\left(x\right)}=4x^3+x^2-2x+7\)

\(b,M_{\left(x\right)}=P_{\left(x\right)}+Q_{\left(x\right)}\\ M_{\left(x\right)}=4x^3+x^2-2x+7-4x^3-x^2+4x-3\\ M_{\left(x\right)}=2x+4\)

\(N_{\left(x\right)}=4x^3+x^2-2x+7+4x^2+x^2-4x+3\\ N_{\left(x\right)}=8x^3+2x^2-6x+10\)

\(c,M_{\left(x\right)}=0\\ \Rightarrow2x+4=0\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\)

a)\(P\left(x\right)=4x^3+x^2-2x+7\)

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

b)\(M\left(x\right)=4x^3+x^2-2x+7-4x^3-x^2-4x+3\)

\(M\left(x\right)=-6x+10\)

\(N\left(x\right)=4x^3+x^2-2x+7+4x^3+x^2+4x-3\)

\(N\left(x\right)=8x^3+2x^2+2x+4\)

c) cho M(x) = 0

\(=>-6x+10=0\)

\(-6x=-10\Rightarrow x=-\dfrac{10}{-6}=\dfrac{5}{3}\)

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

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

1: \(A=5x^5-5x^3+7x^2-2x+4\)

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

2: \(A\left(x\right)+B\left(x\right)=5x^5-5x^3+7x^2-2x+4-5x^6+2x^4+4x^3+4x^2-4x-1\)

\(=-5x^6+5x^5+2x^4-x^3+11x^2-6x+3\)

\(A\left(x\right)-B\left(x\right)\)

\(=5x^5-5x^3+7x^2-2x+4+5x^6-2x^4-4x^3-4x^2+4x+1\)

\(=5x^6+5x^5-2x^4-9x^3+3x^2+2x+5\)

dsssws

a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)

b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)

\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)

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

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

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

1) \(x^2-y^2-2x-2y\)

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

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

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

2) \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)

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

\(=\left(x-y\right)\left(x+5y\right)\)

1) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

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

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

2) 3x² - 3y² - 2(x - y)²

= (3x² - 3y²) - 2(x - y)²

= 3(x² - y²) - 2(x - y)²

= 3(x - y)(x + y) - 2(x - y)²

= (x - y)[3(x + y) - 2(x - y)]

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

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

`x^2-y^2 -2x-2y`

`= (x^2-y^2) -(2x+2y)`

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

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

__

`3x^2 -3y^2 -2(x-y)^2`

`= 3(x^2 -y^2) - 2(x-y)^2`

`=3(x-y)(x+y) -2(x-y)^2`

`= (x-y) (3x+3y -2x+2y)`

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

a, \(P\left(x\right)=5x^2-3x+7\)

\(Q\left(x\right)=-5x^3-x^2+4x-5\)

b, Thay x = 1 vào Q(x) ta được 

-5 - 1 + 4 - 5 = -7 

c, \(Q\left(x\right)+P\left(x\right)=-5x^3+4x^2+x+2\)

\(Q\left(x\right)-P\left(x\right)=-5x^3-6x^2+7x-12\)

\(-5x^3+9x^2+x=0\Leftrightarrow x\left(-5x^2+9x+1\right)=0\Leftrightarrow x=0;x=\dfrac{9\pm\sqrt{101}}{10}\)

Trước hết, ta rút gọn các đa thức:

- Q(x) = 4x3 – 2x + 5x2 - 2x3 + 1 - 2x3

Q(x) = (4x3- 2x3- 2x3) – 2x + 5x2 + 1

Q(x) = 0 – 2x + 5x2 + 1

Q(x) = – 2x + 5x2 + 1

- R(x) = - x2 + 2x4 + 2x - 3x4 – 10 + x4

R(x) = - x2 + (2x4- 3x4+ x4) + 2x – 10

R(x) = - x2 + 0 + 2x – 10

R(x) = - x2 + 2x – 10

Sắp xếp các hạng tử của đa thức sau theo lũy thừa giảm dần của biến ta có:

Q(x) = 5x2 – 2x + 1

R(x) = - x2 + 2x – 10