Bài 6:
a: \(x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
b: \(-x^2+6x-11\)
\(=-\left(x^2-6x+11\right)\)
\(=-\left(x^2-6x+9+2\right)=-\left(x-3\right)^2-2\le-2\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
Bài 5:
a: \(a^2\left(a+1\right)+2a\left(a+1\right)\)
\(=\left(a+1\right)\left(a^2+2a\right)\)
=a(a+1)(a+2)⋮6
b: a(2a-3)-2a(a+1)
\(=2a^2-3a-2a^2-2a\)
=-5a⋮5
c: \(x^2+2x+2\)
\(=x^2+2x+1+1\)
\(=\left(x+1\right)^2+1\ge1>0\forall x\)
Bài 3:
a: \(3n^3+10n^2-5\) ⋮3n+1
=>\(3n^3+n^2+9n^2+3n-3n-1-4\) ⋮3n+1
=>-4⋮3n+1
=>3n+1∈{1;-1;2;-2;4;-4}
=>3n∈{0;-2;1;-3;3;-5}
mà 3n⋮3
nên 3n∈{0;-3;3}
=>n∈{0;-1;1}
b: \(10n^2+n-10\) ⋮n-1
=>\(10n^2-10n+11n-11+1\) ⋮n-1
=>1⋮n-1
=>n-1∈{1;-1}
=>n∈{2;0}
