\(\frac{3x^3+9x^2-x-5}{x+3}=\left(3x^2-1\right)-\frac{2}{x+3}\)là số nguyên khi x+3 là ước của 2, vậy x=-5;-4;-2;-1

thực hiện phép chia đa thức ta có:

\(x^3-5x^2+9x-2=\left(x^2-2x+3\right)\left(x-3\right)+7\)

=> \(A=x^2-2x+3+\frac{7}{x-3}\)

Với x thuộc Z để A thuộc Z thì \(\frac{7}{x-3}\in Z\)<=> \(7⋮\left(x-3\right)\)<=> x-3 thuộc Ư(7). Em tự làm tiếp nhé!

a: ĐKXĐ: x∉{0;3;-3}

\(\frac{9}{x^3-9x}+\frac{1}{x+3}\)

\(=\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\)

\(=\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

\(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\)

\(=\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\)

\(=\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}=\frac{3x-9-x^2}{3x\cdot\left(x+3\right)}=\frac{-\left(x^2-3x+9\right)}{3x\left(x+3\right)}\)

Ta có: \(A=\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}:\frac{-\left(x^2-3x+9\right)}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\frac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}=\frac{-3}{x-3}\)

b: A=2

=>\(-\frac{3}{x-3}=2\)

=>\(x-3=-\frac32\)

=>\(x=-\frac32+3=\frac32\) (nhận)

c: A nguyên

=>-3⋮x-3

=>x-3∈{1;-1;3;-3}

=>x∈{4;2;6;0}

Kết hợp ĐKXĐ, ta được: x∈{4;2;6}

\(a,\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

vậy_____

Vì (x+1).(x-2)=0

suy ra 1 trong 2 so =0

tu giai not

\(\Leftrightarrow2\sqrt{x}-4⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3\in\left\{1;-1\right\}\)

hay \(x\in\left\{4;1\right\}\)