Ta có: \(\left(x^2-3\right).\left(x^2-36\right)\le0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x^2-3\ge0\\x^2-36\le0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2\ge3\\x^2\le36\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\ge\sqrt{3}ho\text{ặc}x\le-\sqrt{3}\\x\le6ho\text{ặc}x\ge-6\end{cases}}}\)
\(\orbr{\begin{cases}x^2-3\le0\\x^2-36\ge0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2\le3\\x^2\ge36\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\le\sqrt{3}ho\text{ặc}x\ge-\sqrt{3}\\x\ge6ho\text{ặc}x\le-6\end{cases}}}\)
KL:................................................................................................................
( x^2 - 3 )( x^2 - 36 ) \(\le0\)
TH1 : ( x^2 - 3 )( x^2 - 36 ) = 0
=> x^2 - 3 = 0 hoac x^2 - 36 = 0
=> x^2 = 3 hoac x^2 = 36
=> x = \(\sqrt{3}\)hoac bang 6 , -6
TH2 : ( x^2 - 3 )( x^2 - 36 ) < 0
=> x^2 - 3 am va x^2 - 36 duong hoac x^2 - 36 am va x^2 - 3 duong
TH x^2 - 3 am ( 1 ) va x^2 - 36 duong ( 2 )
Xet ( 1 ) thi :
=> x^2 < 2
=> x thuoc 1,0,-1
Nhung de x^2 - 36 duong ( 2 ) thi IxI > 6
Ma 1,0,-1 deu < 6
=> x \(\varnothing\)
TH x^2 - 36 am ( 1 ) va x^2 - 3 duong ( 2 )
Xet ( 1 ) thi :
I x I < 6
=> x \(\in\left\{5,4,3,2,1,0,-1,-2,-3,-4,-5\right\}\)
Xet ( 2 ) thi :
I x I > 2
=> x thuoc { 5,4,3,-3,-4,-5 }
Vay x \(\in\left\{\sqrt{3},6,5,4,3,-3,-4,-5,-6\right\}\)
Đặt x2 = n (x là số chính phương)
Ta có: \(\left(x^2-3\right)\left(x^2-36\right)\le0\Leftrightarrow\left(n-3\right)\left(n-36\right)\le0\)
\(\Leftrightarrow\hept{\begin{cases}n-3\ge0\\n-36\le0\end{cases}}\) hoặc \(\hept{\begin{cases}n-3\le0\\n-36\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n\ge3\\n\le36\end{cases}}\) hoặc \(\hept{\begin{cases}n\le3\\n\ge36\end{cases}}\) (loại)
Vậy \(3\le n\le36\)
Các số chính phương x thỏa mãn: \(3\le n\le36\) là: 4;9;16;25;36
Tức là: n = {2;-2;3;-3;4;-4;5;-5;6;-6}
Vậy có tất cả 10 số nguyên x thỏa mãn