\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+x^2+6x+9+1978\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+\left(x+3\right)^2+1978\)

\(=\left(x-y+1\right)^2+\left(x+3\right)^2+1978\ge1978\)

\(A_{min}=1978\) khi \(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

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 \(A_{min}=-\frac{19}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

B2: \(\Rightarrow16x^2-8x-\left(16x^2-8x+1\right)-13\Rightarrow16x^2-8x-16x^2+8x-1-13\Rightarrow-14\)

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

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

\(\Leftrightarrow A=2\left(x^2-2.2.x+4+1\right)\)

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

Dấu " = " xảy ra khi và chỉ khi

\(2\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy min A = 2 <=> x = 2

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

mà \(\left(2x-1\right)^4\ge0,\forall x\)

\(\Rightarrow A=\left(2x-1\right)^4+3\ge0+3=3\)

\(\Rightarrow GTNN\left(A\right)=3\left(x=\dfrac{1}{2}\right)\)

\(B=-\left(8x-\dfrac{4}{5}\right)^6+1\)

mà \(-\left(8x-\dfrac{4}{5}\right)^6\le0,\forall x\)

\(\Rightarrow B=-\left(8x-\dfrac{4}{5}\right)^6+1\le0+1=1\)

\(\Rightarrow GTLN\left(B\right)=1\left(x=\dfrac{1}{10}\right)\)

Giups mik vs

b) B=-3(x^2-3x+9/4)+27/4=-3(x-3/2)^2+27/4 <=27/4. Vậy MaxB=27/4, dấu "=" xảy ra <=> x-3/2=0 <=> x=3/2

a, Ta có :  \(A=2x^2-8x-10=2\left(x^2-4x-5\right)\)

\(=2\left(x^2-4x+4-9\right)=3\left(x-2\right)^2-18\ge-18\)

Dấu ''='' xảy ra <=> x = 2 

Vậy GTNN A là -18 <=> x = 2 

b, Ta có :  \(B=-3x^2+9x=-3\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)\)

\(=-3\left(x-\frac{3}{2}\right)^2+\frac{18}{4}\ge\frac{18}{4}\)

Dấu ''='' xảy ra <=> x = 3/2 

Vậy GTNN B là 18/4 <=> x = 3/2 

với GTLN thì ngược dấu lại làm tương tự như trên.

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

Dấu \("="\Leftrightarrow x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

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

Dấu \("="\Leftrightarrow x=1\)

\(D=\dfrac{1}{-\left(x^2+2x+1\right)+6}=\dfrac{1}{-\left(x+1\right)^2+6}\ge\dfrac{1}{6}\)

Dấu \("="\Leftrightarrow x=-1\)

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

\(A_{min}=-7\) khi \(x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{3}{2}\)

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

\(C_{min}=-4\) khi \(x=1\)

Biểu thức D không tồn tại cả max lẫn min

1.

$A=2x^2-8x+1=2(x^2-4x+4)-7=2(x-2)^2-7$

Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow A\geq 2.0-7=-7$

Vậy $A_{\min}=-7$ khi $x-2=0\Leftrightarrow x=2$

2.

$B=x^2+3x+2=(x^2+3x+1,5^2)-0,25=(x+1,5)^2-0,25\geq 0-0,25=-0,25$

Vậy $B_{\min}=-0,25$ khi $x=-1,5$

3.

$C=4x^2-8x=(4x^2-8x+4)-4=(2x-2)^2-4\geq 0-4=-4$

Vậy $C_{\min}=-4$ khi $2x-2=0\Leftrightarrow x=1$

4. Để $D_{\min}$ thì $5-x^2-2x$ là số thực âm lớn nhất

Mà không tồn tại số thực âm lớn nhất nên không tồn tại $x$ để $D_{\min}$

A\(=2x^2-8x+1\)

=2x(x-4)+1≥1

Min A=1 ⇔x=4

B=\(x^2+3x+2\)

\(=\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{1}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)≥\(-\dfrac{1}{4}\)

Min B=-1/4⇔x=-3/2

C=\(4x^2-8x\)

=\(\left(\left(2x\right)^2-2x.4+16\right)-16\)

=(2x-4)^2 -16≥-16

Min C=-16 ⇔x=2

D=\(\dfrac{1}{-\left(x^2-2x+1\right)+6}\)

=\(\dfrac{1}{-\left(x-1\right)^2+6}\)≥\(\dfrac{1}{6}\)

Min D=1/6 ⇔x=1