2.a) (ko phân tích được, bạn coi lại nhé)

b) phần còn lại của chứng minh là gì thế bạn?

Bài 2:

a: \(A=x^2\left(x-1\right)^2+2x^2-4x-1\)

\(=x^2\left(x^2-2x+1\right)+2x^2-4x-1\)

\(=x^4-2x^3+x^2+2x^2-4x-1\)

\(=x^4-2x^3+3x^2-4x-1\)

\(=\left(x^4-2x^3+x^2\right)+2\left(x^2-2x+1\right)-3\)

\(=\left(x^2-x\right)^2+2\left(x-1\right)^2-3\ge-3\forall x\)

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

\(x^3-9x^2+26x-24\)

\(=x^3-4x^2-5x^2+20x+6x-24\)

\(=\left(x-4\right)\left(x^2-5x+6\right)\)

\(=\left(x-4\right)\left(x-2\right)\left(x-3\right)\)

Bài 2:

a: \(A=x^2\left(x-1\right)^2+2x^2-4x-1\)

\(=x^2\left(x^2-2x+1\right)+2x^2-4x-1\)

\(=x^4-2x^3+x^2+2x^2-4x-1\)

\(=x^4-2x^3+3x^2-4x-1\)

\(=\left(x^4-2x^3+x^2\right)+2\left(x^2-2x+1\right)-3\)

\(=\left(x^2-x\right)^2+2\left(x-1\right)^2-3\ge-3\forall x\)

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

Bài 2:

a: \(A=x^2\left(x-1\right)^2+2x^2-4x-1\)

\(=x^2\left(x^2-2x+1\right)+2x^2-4x-1\)

\(=x^4-2x^3+x^2+2x^2-4x-1\)

\(=x^4-2x^3+3x^2-4x-1\)

\(=\left(x^4-2x^3+x^2\right)+2\left(x^2-2x+1\right)-3\)

\(=\left(x^2-x\right)^2+2\left(x-1\right)^2-3\ge-3\forall x\)

Dấu '=' xảy ra khi \(\begin{cases}x^2-x=0\\ x-1=0\end{cases}\Rightarrow x=1\)

b: \(B=\left(x-5\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+2022\)

\(=\left(x-5\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)+2022\)

\(=\left(x^2-x-20\right)\left(x^2-x-6\right)+2022\)

\(=\left(x^2-x-6\right)^2-14\left(x^2-x-6\right)+49+1973=\left(x^2-x-6+7\right)^2+1973\)

\(=\left(x^2-x+1\right)^2+1973\)

Ta có: \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34\forall x\)

=>\(\left(x^2-x+1\right)^2\ge\frac{9}{16}\forall x\)

=>\(\left(x^2-x+1\right)^2+1973\ge\frac{9}{16}+1973\forall x\)

=>B>=31577/16∀x

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

a)9x²+12+7

      Sai đề trầm trọng

b)x²-26x+180

         Ta có:x²-26x+180=x²+2.13x+13²+11

                                     =(x+13)²+11

     Vì (x+13)²\(\ge\)0

                             Suy ra:(x+13)²+11\(\ge\)11

Dấu = xảy ra khi x+13=0

                           x=-13

Vậy Min B=11 khi x=-13

a) \(9x^2+12x+7=\left(9x^2+12x+4\right)+3=\left(3x+2\right)^2+3\ge3\)

Min = 3 <=> x = -2/3

b) \(x^2-26x+180=\left(x^2-26x+169\right)+11=\left(x-13\right)^2+11\ge11\)

Min = 11 <=> x = 13

\(A=\frac{5x^2-26x+41}{\left(x-2\right)^2}=\frac{4\left(x^2-4x+4\right)+\left(x^2-10x+25\right)}{\left(x-2\right)^2}=4+\frac{\left(x-5\right)^2}{\left(x-2\right)^2}\ge4\forall x\)

Dấu "=" xảy ra khi \(x-5=0\Rightarrow x=5\)

Vậy GTNN của A là 4 khi x = 5

a) \(\frac{5x^2-20x+20-6x+21}{\left(x-2\right)^2}=\frac{5\left(x^2-4x+4\right)-6\left(x-2\right)+9}{\left(x-2\right)^2}\)

=\(\frac{5\left(x-2\right)^2-6\left(x-2\right)+9}{\left(x-2\right)^2}=5-\frac{6}{\left(x-2\right)}+\frac{9}{\left(x-2\right)^2}=\left(\frac{3}{x-2}-1\right)^2+4\ge4\)

'=' xảy ra \(\Leftrightarrow\frac{3}{x-2}-1=0\Leftrightarrow x=5\)

Vậy ...