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

chị thấy câu B hơi rối

Lời giải:

a.

$A+B=(5x^2-7x+2)+(4x^2+3x-1)=9x^2-4x+1$
$A-B=(5x^2-7x+2)-(4x^2+3x-1)=x^2-10x+3$

b. 

$A(x)=2x^2-x+m=x(2x-5)+4x+m=x(2x-5)+2(2x-5)+m+10$

$=B(x)(x+2)+m+10$

Để $A(x)\vdots B(x)$ thì $m+10=0\Leftrightarrow m=-10$

a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)

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

f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

=\(3x^5-10x^4-13\)

b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)

=\(x^4+9x^3-11x^2+7x-2\)

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

=\(9x^4+5x^3-x^2-x-12\)

a ) 

\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)

\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)

P(x) chia hết cho x-2 cần P(2)-0 nên thay x=2 vào P(x) được: P(x)=2^4-5.2^3-4.x^2+3.2+m=m-34=0 =>m=34

tương tự tìm n=-40

tại sao P(x) muốn chia hết cho x-2 thì P(2) phải bằng 0

c: \(P\left(-1\right)=-3-5-4+2+6+4=0\)

Vậy: x=-1 là nghiệm của P(x)

\(Q\left(-1\right)=4+1+3+2-7+1=4< >0\)

=>x=-1 không là nghiệm của Q(x)

1:

a: f(x)=2x^4+2x^3+2x^2+5x+6

g(x)=x^4-2x^3-x^2-5x+3

c: h(x)=2x^4+2x^3+2x^2+5x+6+x^4-2x^3-x^2-5x+3=3x^4+x^2+9

K(x)=f(x)-2g(x)-4x^2

=2x^4+2x^3+2x^2+5x+6-2x^4+4x^3+2x^2+10x-6-4x^2

=6x^3+15x

c: K(x)=0

=>6x^3+15x=0

=>3x(2x^2+5)=0

=>x=0

d: H(x)=3x^4+x^2+9>=9

Dấu = xảy ra khi x=0

Bài 2:

M = \(\frac{9x+5}{3x-1}\)

M ∈ Z ⇔ (9x + 5) ⋮ (3x -1)

[3.(3x - 1) + 8] ⋮ (3x -1)

8 ⋮ (3x -1)

(3x - 1) ∈ Ư(8) = {-8; - 4; -2; -1; 1; 2; 4; 8}

x ∈ {-7/3; -1; -1/3; 0; 2/3; 1; 5/3; 3}

Vì x ∈ Z Vậy x ∈ {0; 1; 3}