A = 2x^2 - 8 x + 10 + (y-3)^4

A = (2x^2 - 8x + 8) + (y-3)^4 + 2

A = 2.(x^2 - 4x + 4) + (y-3)^4 + 2

A = 2.(x^2-2)^2 + (y-3)^4 + 2 >= 2.

Dấu "=" xảy ra <=> x^2 - 2 = 0 và y - 3 = 0

<=> x = \(\pm\sqrt{2}\)và y = 3.

Vậy Min A = 2 <=> x = \(\pm\sqrt{2}\)và y = 3

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

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

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

\(A_{min}=1978\) 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 = 2x² - 8x + 14

A = 2x² - 4x - 4x + 8 + 6

A = 2x.(x - 2) - 4.(x - 2) + 6

A = (x - 2).(2x - 4) + 6

A = 2.(x - 2)² + 6 \(\ge6\) với mọi x

Dấu "=" xảy ra khi (x - 2)² = 0

=> x - 2 = 0

=> x = 2

Vậy A_Min = 6 khi và chỉ khi x = 2 

A= 2x²-8x+14

=2(x²-4x+7)

=2(x²-4x+4)+6

=2(x-2)²+6\(\ge\)6

Dấu = khi x-2=0 <=>x=2

Vậy MinA=6 khi x=2

\(A=2x^2-8x+14\)

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

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

Ta có : \(2\left(x-2\right)^2\ge0;\forall x\inℝ\)

Suy ra \(2\left(x-2\right)^2+6\ge6\)

Dấu = xảy ra \(< =>2\left(x-2\right)^2=0\)

\(< =>x-2=0< =>x=2\)

Vậy Min A = 6 khi x = 2

=2xx+2x+6x+1

=2x(x+1)+6x+1=2x(x+1+3x)+1≥1

dấu = xảy ra khi 2x(x+1+3x)=0 còn lại bạn tự xử nhé :)

bài này mình ko chắc có đúng ko nên phải nghiên cứu trước  rồi mới làm nha

b: Ta có: \(B=2x^2+8x+1\)

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

\(=2\left(x^2+4x+4-\dfrac{7}{2}\right)\)

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

Dấu '=' xảy ra khi x=-2

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

\(A=\dfrac{2x^2-8x+17}{x^2-2x+1}\left(x\ne1\right)\)

\(\Leftrightarrow A\left(x^2-2x+1\right)=2x^2-8x+17\)

\(\Leftrightarrow Ax^2-2Ax+A=2x^2-8x+17\)

\(\Leftrightarrow x^2\left(A-2\right)-2x\left(A-4\right)+A-17=0\left(1\right)\)

\(A-2=0\Leftrightarrow A=2\Leftrightarrow x=3,75\left(tm\right)\left(2\right)\)

\(A-2\ne0\Leftrightarrow A\ne2\Rightarrow\Delta'\ge0\Leftrightarrow\left(A-4\right)^2-\left(A-17\right)\left(A-2\right)\ge0\Leftrightarrow A\ge\dfrac{18}{11}\Rightarrow A_{min}=\dfrac{18}{11}\Leftrightarrow x=\dfrac{13}{2}\left(tm\right)\left(3\right)\)

\(\left(2\right)và\left(3\right)\Rightarrow A_{min}=\dfrac{18}{11}\Leftrightarrow x=\dfrac{13}{2}\)