giúp mk voi m n

mifh ko giúp dc rùi hihi

\(\left|a^2-3a+1\right|=1\)

=>\(\left[\begin{array}{l}a^2-3a+1=1\\ a^2-3a+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2-3a=0\\ a^2-3a+2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}a\left(a-3\right)=0\\ \left(a-1\right)\left(a-2\right)=0\end{array}\right.\Rightarrow a\in\left\lbrace0;1;2;3\right\rbrace\)

ĐKXĐ: a<>2

=>a∈{0;1;3}

\(A=\frac{2a^3-12a^2+17a-2}{a-2}\)

\(=\frac{2a^3-4a^2-8a^2+16a+a-2}{a-2}=\frac{2a^2\left(a-2\right)-8a\left(a-2\right)+\left(a-2\right)}{a-2}\)

\(=2a^2-8a+1\)

Khi a=0 thì \(A=2a^2-8a+1=2\cdot0^2-8\cdot0+1=1\)

Khi a=1 thì \(A=2a^2-8a+1=2\cdot1^2-8\cdot1+1=2-8+1=3-8=-5\)

Khi a=3 thì \(A=2a^2-8a+1=2\cdot3^2-8\cdot3+1=18-24+1=19-24=-5\)

Chọn D.

Tại a = -9 ta được:

= 3√-(-9) - |3 + 2(-9)|

= 3√32 - |3 - 18|

= 3.3 - |-15| = 9 - 15 = -6

Ta có A = 12   –   ( 2 a   –   3 ) 2   ≤   12 ∀ a  Þ A_max = 12 Û a = 3 2 .

P=4a²+4ab+4b²-12a-12b+12

=[(4a²-12a+9)+2b(2a-3)+b²]+3b²-6b+12

=(2a+b-3)²+3(b-1)²+9

Dấu "=" xảy ra khi b-1=0=> b=1

                        và 2a+b-3=0 => 2a+1-3=0=> a=1

Vậy MinP = 9 <=> a=b=1