F(\(x\)) = 3\(x^3\) - 2\(x^2\) + 1

F(-2) = 3(-2)³ - 2.(-2)² + 1

F(-2) = -24 - 8 + 1

F(-2) = -32 + 1

F(-2) = -31

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

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

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

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

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

\(=2x^7+x^3+x^2+6x\)

a: \(A\left(x\right)=x-2x^2+3x^5+x^4+x+x^2\)

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

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

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

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

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

b: B(x)=A(x)+M(x)

=>M(x)=B(x)-A(x)

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

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

bậc là 5

Hệ số cao nhất là -6

Lời giải:

$|x-2023|=x-2023$

$\Rightarrow x-2023\geq 0$

$\Rightarrow x\geq 2023$.

x thỏa mãn tất cả các GT .

Lời giải:

$2xy=x+y$

$\Rightarrow 2xy-x-y=0$

$\Rightarrow x(2y-1)-y=0$

$\Rightarrow 2x(2y-1)-2y=0$

$\Rightarrow 2x(2y-1)-(2y-1)=1$

$\Rightarrow (2x-1)(2y-1)=1$

Do $x,y$ nguyên nên $2x-1,2y-1$ nguyên. Mà tích của chúng bằng $1$ nên ta xét các TH sau:

TH1: $2x-1=1, 2y-1=1\Rightarrow x=y=1$

TH2: $2x-1=-1, 2y-1=-1\Rightarrow x=y=0$

Vậy........

Lời giải:

a.

$(4x^4+3x^3):(-x^3)+(15x^2+6x):3x=0$

$\Rightarrow -4x-3+(5x+2)=0$

$\Rightarrow -4x-3+5x+2=0$

$\Rightarrow x-1=0$

$\Rightarrow x=1$
b.

$(x^2-\frac{1}{2}x):(2x)-(3x-1)^2:(3x-1)=0$

$\Rightarrow \frac{x}{2}-\frac{1}{4}-(3x-1)=0$

$\Rightarrow \frac{x}{2}-\frac{1}{4}-3x+1=0$

$\Rightarrow \frac{-5}{2}x+\frac{3}{4}=0$

$\Rightarrow \frac{-5}{2}x=\frac{-3}{4}$

$\Rightarrow x=\frac{3}{10}$

a: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)

=>\(-4x-3+5x+2=0\)

=>x-1=0

=>x=1

b: \(\left(x^2-\dfrac{1}{2}x\right):2x-\left(3x-1\right)^2:\left(3x-1\right)=0\)

=>\(\dfrac{1}{2}x-\dfrac{1}{4}-\left(3x-1\right)=0\)

=>\(\dfrac{1}{2}x-\dfrac{1}{4}-3x+1=0\)

=>\(-\dfrac{5}{2}x=\dfrac{1}{4}-1=-\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}:\dfrac{5}{2}=\dfrac{3}{4}\cdot\dfrac{2}{5}=\dfrac{6}{20}=\dfrac{3}{10}\)