Question

Tim x \(\in z\);

\(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)<0\)

Answer 1

Ta có: \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)<0\)

=>\(\left[\left(x^2-1\right)\left(x^2-7\right)\right].\left[\left(x^2-4\right)\left(x^2-10\right)\right]<0\)

=>\(\left[\left(x^2-4+3\right)\left(x^2-4-3\right)\right].\left[\left(x^2-7+3\right)\left(x^2-7-3\right)\right]<0\)

=>\(\left[\left(x^2-4\right)^2-3^2\right].\left[\left(x^2-7\right)^2-3^2\right]<0\)

=>\(\left[\left(x^2-4\right)^2-9\right].\left[\left(x^2-7\right)^2-9\right]<0\)

=>(x²-4)-9 và (x²-7)-9 khác dấu

Vì \(\left(x^2-4\right)^2-9>\left(x^2-7\right)^2-9\)

=>\(\left(x^2-4\right)^2-9>0=>\left(x^2-4\right)^2>9=>x^2-4>3=>x^2>7=>x>2\)

Và \(\left(x^2-7\right)^2-9<0=>\left(x^2-7\right)^2<9=>x^2-7<3=>x^2<10=>x<4\)

=>2<x<4

mà \(x\in Z\)

=>x=3

Vậy x=3