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

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

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

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

Dấu " = " xảy ra khi và chỉ khi

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

Vậy min A = 2 <=> x = 2

\(M=2x^2-8x+\sqrt{x^2-4x+5}+6\)

\(=2\left(x^2-4x+5\right)+\sqrt{x^2-4x+5}-4\)

Đặt \(\sqrt{x^2-4x+5}=t\)

Ta thấy \(x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x+2\right)^2+1\ge1\)

Vậy nên \(\sqrt{x^2-4x+5}\ge1\Rightarrow t\ge1\)

Khi đó \(M=2t^2+t-4=2\left(t^2+\frac{1}{2}t-2\right)=2\left[\left(t^2+2.t.\frac{1}{4}+\frac{1}{16}\right)-\frac{33}{16}\right]\)

\(=2\left[\left(t+\frac{1}{4}\right)^2-\frac{33}{16}\right]=2\left(t+\frac{1}{4}\right)^2-\frac{33}{8}\)

Do \(t\ge1,\left(t+\frac{1}{4}\right)^2\ge\frac{25}{16}\)

Vậy thì \(M\ge2.\frac{25}{16}-\frac{33}{8}=-1\)

Vậy \(minM=-1\) khi t = 1 

hay \(\sqrt{x^2-4x+5}=0\Rightarrow x^2-4x+5=2\Rightarrow x^2-4x+4=0\Rightarrow x=2\)

Ta có:

\(C=2x^2+3y^2+4xy-8x-2y+18\)

\(C=2\left(x^2+2xy+y^2\right)+y^2-8x-2y+18\)

\(C=2[\left(x+y\right)^2-4\left(x+y\right)+4]+\left(y^2+6y+9\right)+1\)

\(C=2\left(x+y-2\right)^2+\left(y+3\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow x+y=2\)và \(y=-3\)

Hay x = 5 , y = -3

\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

\(b,B=4x^2+4x-2\)

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

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

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

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

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

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

\(\Leftrightarrow x=-\frac{1}{2}\)

A=0

B=4/3

C=o

giải thích chi tiết giúp mình được không bạn ;-;