a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

$=x^2(x-5)-(x-5)=0$

$=(x-5)(x^2-1)=0$

$=(x-5)(x-1)(x+1)=0$

$\Rightarrow x=5,\ x=1,\ x=-1$

$3x^4-9x^3+9x^2-27x=0$

$3x(x^3-3x^2+3x-9)=0$

$3x[x^2(x-3)+3(x-3)]=0$

$3x(x-3)(x^2+3)=0$

Vì $x^2+3>0$ nên:

$\Rightarrow x=0$ hoặc $x=3$

1. A

2. D

3. A

4. A

4A

3A

2D

1D

a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a)\(\left(x2+7\right).\left(x2-49\right)< 0\)

\(\left(x2+7\right).\left(x2-49\right)< 0\) chứng tỏ hai vế \(\left(x2+7\right)\) và \(\left(x2-49\right)\) khác dấu nhau .

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)

Vì \(\left(x2+7\right)\) > \(\left(x2-49\right)\)

Nên ta có:

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\left(x+7\right)=0\\\left(x-49\right)=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=-7\\x=49\end{matrix}\right.\)

Vậy hai số nguyên đó là -7 và 49 .

Còn phần còn lại bạn làm tương tự nhé !