2. a. \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18\)

\(=2\left(x-2\right)^2-18\)

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy minA = - 18 <=> x = 2

b. \(B=9x-3x^2=-3\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)

\(=-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\le\frac{27}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Vậy maxB = 27/4 <=> x = 3/2

Sửa đề:x³-3x²-4x+12

a,x³-3x²-4x+12

=(x³-3x²)-(4x+12)

=x²(x-3)-4(x-3)

=(x²-4)(x-3)

b,x⁴- 5x² +4

x⁴-4x²-x²+4

(x⁴-x²)-(4x²+4)

x²(x²-1)-4(x²-1)

(x²-4)(x²-1)

a. Ta có:

\(A=2x^2-8x+10\\ =2\left(x^2-4x+5\right)=2\left[\left(x^2-2.x.2+4\right)+1\right]\\ =2\left[\left(x-2\right)^2+1\right]\\ =2\left(x-2\right)^2+2\)

Vì \(\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2\ge0\forall x\\ \Leftrightarrow2\left(x-2\right)^2+1\ge1\forall x\)

Dấu = xảy ra khi: \(2\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(MinA=1\Leftrightarrow x=2\)

GTLN

\(A=2x^2-8x-10\)

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

\(A=2\left(x^2-2.x.2+2^2-2^2-5\right)\)

\(A=2\left[\left(x^2-4x+2^2\right)-4-5\right]\)

\(A=2\left(x-2\right)^2-9\)

suy ra \(\left(x-2\right)^2\ge-9\)

=> Min A = -9 khi \(\left(x-2\right)^2=0\Leftrightarrow x-2=0hayx=2\)

Vậy Min A = (-9) khi x =2

\(A=x^2+x\) . Có: \(x^2\ge x\Rightarrow x^2+x\ge0\)

Dấu '=' xảy ra khi: \(x^2+x=0\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy: \(Min_A=0\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

\(B=4x-12x+10\)

\(B=-8x+10\)

\(B=10-8x\)

Xét: \(x< 0\Rightarrow10-8x\ge10\)

Dấu '=' xảy ra khi: \(8x=0\Rightarrow x=0\)

Xét: \(x>0\Rightarrow10-8x\le10\)

Dấu '=' xảy ra khi: \(8x=0\Rightarrow x=0\)

Vậy: Khi x<0. \(Min_B=10\) tại \(x=0\)

Khi: x>0.  \(Max_B=10\)tại \(x=0\)

K chắc

a) \(A=2x^2\)\(+\)\(10\)\(-\)\(1\)

\(=2\left(x^2+5x-\frac{1}{2}\right)\)

\(=2\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)

\(=2\left(x+\frac{5}{2}\right)^2\)\(=\frac{27}{2}\)> hoặc = \(\frac{-27}{2}\)\(=-13,5\)

Dấu bằng xảy ra  \(\Leftrightarrow\)\(x+\frac{5}{2}=0\)

                                    \(x=\frac{-5}{2}=-2,5\)

Vậy GTLN của A bằng -13,5 khi x = -2,5

b)  \(B=3x-2x^2\)

\(=\)\(-2\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}-\frac{9}{16}\right)\)

\(=-2\left[\left(x-\frac{3}{4}\right)^2-\frac{9}{16}\right]\)

\(=-2\left(x-0,75\right)^2\)\(+\)\(\frac{9}{8}\)< hoặc = \(\frac{9}{8}\)\(=\)\(1,125\)

Dấu bằng xảy ra  \(\Leftrightarrow\)\(x-0,75=0\)

                                    \(x=0,75\)

Vậy GTLN của B bằng 1,125 khi x = 0,75

kjkkm

a) Ta có : \(A=-6x+x^2+11\)

\(\Rightarrow A=\left(x^2-6x+9\right)+2\)

\(\Rightarrow A=\left(x-3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy \(minA=2\Leftrightarrow x=3\)

b) \(B=-1+2x^x+10x\)

\(\Rightarrow\)Tớ đang thắc mắc cái chỗ 2x^x :)))

làm a)  GTLN = -2( x² -4x +2) + 4 

GTLN =4

A=[2(x^2-8x+22)-1]/(x^2-8x+22)

A=2-1/[(x-4)^2+6]

A nho nhat khi (x-4)^2=0=> x=4

min(A)=2-1/6

B = 4x² + 8x 

= 4( x² + 2x + 1 ) - 4

= 4( x + 1 )² - 4

4( x + 1 )² ≥ 0 ∀ x => 4( x + 1 )² - 4 ≥ -4

Đẳng thức xảy ra <=> x + 1 = 0 => x = -1

=> MinB = -4 <=> x = -1

C = -2x² + 8x - 15

= -2( x² - 4x + 4 ) - 7

= -2( x - 2 )² - 7

-2( x - 2 )² ≤ 0 ∀ x => -2( x - 2 )² - 7 ≤ -7

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxC = -7 <=> x = 2