a, \(\left(x-2\right).\left(x^2+1\right)>0\) \(\Rightarrow x-2>0\) (vì \(x^2+1>0\forall x\inℤ\) )
\(\Rightarrow x>2\)
b, \(\left(x+5\right).\left(2-x\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x>2\\x< -5\end{cases}}\)
a, \(\left(x-2\right).\left(x^2+1\right)>0\) \(\Rightarrow x-2>0\) (vì \(x^2+1>0\forall x\inℤ\) )
\(\Rightarrow x>2\)
b, \(\left(x+5\right).\left(2-x\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x>2\\x< -5\end{cases}}\)
Tim so nguyen x biet
(x - 2)(x2 + 1) > 0
(x + 5)(2 - x) < 0
tim so nguyen x biet:(2-x)(5+x)<0
tim so nguyen x, biet (x^2-5).(x^2-25)<0
Tim so nguyen x,y biet:
a. (x+2).(x+5)>0
b.(x-1).(y-2)=12
Tim so nguyen x biet : x+(x+1)+(x+2)+(x+3)+........+23+24=0
Tim cac so nguyen x biet:
a,(x^2--5)(x^2-25)<0
b,(x-2)(x^2+1)=0
c,(x+3)(x^2+9)<0
d,(x-1)92x^2-8)^2=3
1,tim cac so nguyen x,y biet: -2/x=y/3 va x<0<y
2, Tim cac so nguyen x,y biet:x-3/y-2=3/2 va x-y=4
bai 1 tim so nguyen duong nho nhat co 3 chu so , so nguyen a lon nhat co 2 chu so
bai 2: tim x thuoc Z biet
a, / x/ = -5
b, /x/= <7
c, /x/ = 4
d, /x/=0
tim so nguyen x biet
c)(x - 2)(x + 1)<0
d)(x - 1)((x2 + 4) < 0