a: |4x-1|=1

=>\(\left[\begin{array}{l}4x-1=1\\ 4x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=2\\ 4x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=0\end{array}\right.\)

Thay x=1/2 vào A(x), ta được:

\(A\left(\frac12\right)=\left(\frac12\right)^4-4\cdot\left(\frac12\right)^3+2\cdot\left(\frac12\right)^2-5\cdot\frac12+6\)

\(=\frac{1}{16}-4\cdot\frac18+2\cdot\frac14-\frac52+6=\frac{1}{16}-\frac12+\frac12-\frac52+6\)

\(=\frac{1}{16}-\frac{40}{16}+\frac{96}{16}=\frac{97-40}{16}=\frac{57}{16}\)

Thay x=0 vào A(x), ta được:

\(A\left(0\right)=0^4-4\cdot0^3+2\cdot0^2-5\cdot0+6=6\)

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

=>A(x)-B(x)=\(x^4-3x^3+\left(3x^2-x^2-2x^2\right)-x+6\)

=>A(x)-B(x)=\(x^4-3x^3-x+6\)

=>\(B\left(x\right)=A\left(x\right)-\left(x^4-3x^3-x+6\right)\)

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

c: Đặt B(x)=0

=>\(-x^3+2x^2-4x=0\)

=>\(x^3-2x^2+4x=0\)

=>\(x\left(x^2-2x+4\right)=0\)

mà \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3>0\forall x\)

nên x=0

1: P(x)=M(x)+N(x)

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

=2x^2-8

2: P(x)=0

=>x^2-4=0

=>x=2 hoặc x=-2

3: Q(x)=M(x)-N(x)

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

=-4x^3+8x+2

a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)

Bậc của P(x) là 3

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

Bậc của Q(x) là 3

b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)

Mình cảm ơn

a) P(x) = 5x³ - 3x + 7 - x

        = 5x³ - 4x + 7

Q(x) = -4x³ + 5x² - 3x + 4x + 3x³ - 4x² + 1

        = -x³ + x² + x + 1

b) M(x) = P(x) + Q(x)

             = ( 5x³ - 4x + 7 ) + ( -x³ + x² + x + 1 )

             = 5x³ - 4x + 7 -x³ + x² + x + 1

             = 4x³ + x² - 3x + 8

N(x) = P(x) - Q(x) 

        = ( 5x³ - 4x + 7 ) - ( -x³ + x² + x + 1 )

        = 5x³ - 4x + 7 + x³ - x² - x - 1

        = 6x³ - x² - 5x + 6

c) M(x) =  4x³ + x² - 3x + 8

M(x) = 0 <=> 4x³ + x² - 3x + 8 = 0

( Bạn xem lại đề nhé chứ lớp 7 chưa học tìm nghiệm đa thức bậc 3 đâu ) 

oke bạn, thank bạn nhaaaaa:)

a)\(P\left(x\right)=M\left(x\right)+N\left(x\right)\)

\(P\left(x\right)=x^4+3x-\dfrac{1}{9}-x+3x^4+2x^2+8x-2x^3+2x^3+\dfrac{2}{3}+4x-4x^4-\dfrac{1}{3}\)

\(P\left(x\right)=2x^2+\dfrac{2}{9}+14x\)

rối lắm luôn

a: \(M\left(x\right)=3x^5-2x^3+x^2-6\)

\(N\left(x\right)=-x^4+3x^3-4x^3-2x^2+\dfrac{1}{3}=-x^4-x^3-2x^2+\dfrac{1}{3}\)

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

\(N\left(x\right)=-3x^4+2x^3-5x^2+7x+5\)

\(P\left(x\right)=M\left(x\right)+N\left(x\right)\)

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

\(=3x+6\)

\(Q\left(x\right)=M\left(x\right)-N\left(x\right)\)

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

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

\(=6x^4-4x^3+10x^2-11x-4\)

\(Câu\text{ }4:\\ Ta\text{ }có:\text{(x^2 – 3x + 2) + (4x^3– x^2+ x – 1)}\\ =x^2-3x+2+4x^3-x^2+x-1\\ =\text{4x}^3+\left(x^2-x^2\right)+\left(-3x+x\right)+\left(2-1\right)\\ =4x^3-2x+1\)

\(Câu\text{ }5:Đặt\text{ }tính\text{ }trừ\text{ }như\text{ }sau:\)

-x^3 -5x + 2 _ 3x + 8 x^3 -8x - 6

Câu 6:

A+B

\(=6x^4-4x^3+x-\frac13+\left(-3x^4-2x^3-5x^2+x+\frac23\right)\)

\(=6x^4-4x^3+x-\frac13-3x^4-2x^3-5x^2+x+\frac23\)

\(=3x^4-6x^3-5x^2+2x+\frac13\)

A-B

\(=6x^4-4x^3+x-\frac13-\left(-3x^4-2x^3-5x^2+x+\frac23\right)\)

\(=6x^4-4x^3+x-\frac13+3x^4+2x^3+5x^2-x-\frac23\)

\(=9x^4-2x^3+5x^2-1\)

Câu 4:

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

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

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

Câu 3:

\(A+B=2x^5+5x^3-2\)

=>\(B+x^4-3x^2-2x+1=2x^5+5x^3-2\)

=>\(B=2x^5+5x^3-2-x^4+3x^2+2x-1=2x^5-x^4+5x^3+3x^2+2x-3\)

\(A-C=x^3\)

=>\(x^4-3x^2-2x+1-C=x^3\)

=>\(C=x^4-3x^2-2x+1-x^3\)

Câu 2: