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90.10n -10n+2 +10n+1-20
= 90.10n - 10n.102 + 10n .10 - 20
=10n . ( 90 - 100 + 10 ) - 20
= 10n .0 -20
= 0 - 20
= -20
B3:\(\Rightarrow90.10^n-10^n.10^2+10^n.10-20\Rightarrow10^n.\left(90-10^2\right)+10^n.10-20\)
\(\Rightarrow10^n.\left(90-100\right)+10^n.10-20\Rightarrow-10.10^n+10^n.10-20\Rightarrow-20\)
\(A=-\left(x^2-x+5\right)=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{19}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\right]\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)
Vậy \(A_{min}=-\frac{19}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
90.10n - 10n+2 + 10n+1 - 20
= 9.10.10n - 10n+1 . 10 + 10n+1 - 20
= 9.10n+1 - 10n+1 . 10 + 10n+1 - 20
= 10n+1.(9 - 10 + 1) - 20
= 10n+1.0 - 20
= 0 - 20
= -20
a: =>\(n+2\in\left\{1;-1;7;-7\right\}\)
=>\(n\in\left\{-1;-3;5;-9\right\}\)
b: =>n-3+4 chia hết cho n-3
=>\(n-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(n\in\left\{4;2;5;1;7;-1\right\}\)
c: =>3n^3+n^2+9n^2-1-4 chia hết cho 3n+1
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)
d: =>10n^2-10n+11n-11+1 chia hết cho n-1
=>\(n-1\in\left\{1;-1\right\}\)
=>\(n\in\left\{2;0\right\}\)
90.10n-10n+2+10n+1-20
=90.10n-10n.102+10n.10-20
=10n.90-10n.100+10n.10-20
=10n.(90-100+10)-20
=10n.0-20
=0-20
=-20
90.10n-10n+2+10n+1-20
=10n+1(9-10+1)-20=10n+1.0-20=0-20=-20