Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

\(A=x-1+\dfrac{9}{x-1}+4\ge2\sqrt{\dfrac{9\left(x-1\right)}{x-1}}+4=10\)

\(A_{min}=10\) khi \(x=4\)

\(A=x+\frac{9}{x-1}+3\Leftrightarrow x-1+\frac{9}{x-1}+3\)

Áp dụng cosi 2 số đầu ta được : 

\(x-1+\frac{9}{x-1}\ge2\sqrt{\left(x-1\right)\frac{9}{x-1}}=6\)

Dễ dàng suy ra : \(A\ge3+6=9\)

Dấu ''='' xảy ra <=> \(x-1=\frac{9}{x-1}\Leftrightarrow\left(x-1\right)^2=9\)

TH1 : \(x-1=3\Leftrightarrow x=4\)( chọn )

TH2 : \(x-1=-3\Leftrightarrow x=-2\)( bỏ vì x > 1 ) theo giả thiết 

Vậy GTNN A là 9 <=> x = 4 

Ta có:A=x²-5x+1=\(\left(x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{25}{4}+1=\left(x-\dfrac{5}{4}\right)^2-\dfrac{21}{4}\)

Vì \(\left(x-\dfrac{5}{4}\right)^2\ge0\)

⇒ \(A\ge-\dfrac{21}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

\(A=x^2-8x+5\)

\(=\left(x^2-8x+16\right)-11\)

\(=\left(x-4\right)^2-11\)

\(=-11+\left(x-4\right)^2\)

Vì \(\left(x-4\right)^2\) ≥ 0

⇒ A ≥ -11

Min A=-11 ⇔\(x-4=0\)

                 ⇔\(x=4\)

\(A=x^2-6x+15=\left(x^2-6x+9\right)+6\)

\(=\left(x-3\right)^2+6\ge6\)

\(minA=6\Leftrightarrow x=3\)

A=x²-2x3+3²+6

A=(x-3)²+6

Vì (x-3)² luôn > hoặc = 0 với mọi x

=> (x-3)²+6 > hoặc = 6

Vậy GTNN = 6 

Dấu "=" xảy ra khi x-3=0

X=3

\(A=x^2-6x+15\)

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

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

\(\Rightarrow A=\left(x-3\right)^2+6\)

Ta có: \(\left(x-3\right)^2+6\ge6\) với mọi x

Dấu ''='' xảy ra khi \(x=3\)

Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)

\(=x^2-6x+9+x^2-22x+121\)

\(=2x^2-28x+130\)

\(=2\left(x^2-14x+49+16\right)\)

\(=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu '=' xảy ra khi x=7

ủa, ko cho x thì sao mak làm:?

Ta có: A=2x²-3x+1=\(2\left(x^2-2.\dfrac{3}{4}+\dfrac{9}{16}\right)-\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)

Vì \(2\left(x-\dfrac{3}{4}\right)^2\ge0\)

 \(\Rightarrow A\ge-\dfrac{1}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)

Vậy,Min \(A=\dfrac{-1}{8}\Leftrightarrow x=\dfrac{3}{4}\)

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+y^2-8y+16-17\\ A=\left(x-y+1\right)^2+\left(y-4\right)^2-16\ge17\)

Vậy \(A_{min}=17\leftrightarrow\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)