Question

Tìm các số nguyên x biết:

a) (x-2).(x+1) = 6

b) (x^2+7).(x^2-49) < 0

c) (x^2-7).(x^2-49) <0

Giúp mình với !!!

Answer 1

a) (x-2).(x+1)=6

(x+1-3).(x+1)=6

(x+1).(1-3)=6

(x+1).-2=6

x+1=6:(-2)

x+1=-3

x=-3-1

x=-4

Answer 2

a) Xét ước

b)

\(\left(x^2+7\right)\left(x^2-49\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+7< 0\Rightarrow x^2< -7\left(KTM\right)\\x^2-49>0\Rightarrow x^2>49\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+7>0\Rightarrow x^2>-7\\x^2-49< 0\Rightarrow x^2< 9=49\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-7< x^2< 49\)

\(\Rightarrow0\le x^2< 49\)

\(\Rightarrow x^2\in\left\{0;1;4;9;16;25;36\right\}\)

\(\Rightarrow x\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5;\pm6\right\}\)

\((x^2-7)(x^2-49)<0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-7< 0\Rightarrow x^2< 7\\x^2-49>0\Rightarrow x^2>49\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-7>0\Rightarrow x^2>7\\x^2-49< 0\Rightarrow x^2< 49\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow7< x^2< 49\)

\(\Rightarrow x^2\in\left\{9;16;25;36\right\}\)

\(\Rightarrow x\in\left\{\pm3;\pm4;\pm5;\pm6\right\}\)