\(A=x^2-10x+32=x^2-10x+25+9=\left(x-5\right)^2+9\)

mà \(\left(x-5\right)^2\ge0\)

\(\Rightarrow\left(x-5\right)^2+9\ge9\)

\(\Rightarrow Min\left(A\right)=9\)

$A=x^2-10x+32=x^2-10x+25+9={\left(x-5\right)}^2+9$

mà ${\left(x-5\right)}^2\ge 0$

$\Rightarrow {\left(x-5\right)}^2+9\ge 9$

$\Rightarrow Min\left(A\right)=9$

2: \(A=x^2-10x+25-34=\left(x-5\right)^2-34\ge-34\forall x\)

Dấu '=' xảu ra khi x=5

\(1,C=x^2+x-3\\ \Rightarrow C=\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{13}{4}\\ \Rightarrow C=\left(x+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\)

dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(C_{min}=-\dfrac{13}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

\(2,A=x^2-10x-9\\ \Rightarrow A=\left(x^2-10x+25\right)-34\\ \Rightarrow A=\left(x-5\right)^2-34\)

dấu "=" xảy ra \(\Leftrightarrow x=5\)

Vậy \(A_{min}=-34\Leftrightarrow x=5\)

Lời giải:

Vì $x=9$ nên $x-9=0$
Ta có:

$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$

$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$

$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$

$=x-10=9-10=-1$

\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(A_{min}=3\) khi \(x=-2\)

\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

\(B_{min}=1\) khi \(x=10\)

\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)

a: =25x^2-10x+25x^2-1-10x=50x^2-20x-1

b: =x^2-12x+32-x^2+12x-32

=0

Câu 1

=> 38-x=0 hoặc x+25=0

TH1

38-x=0

x=38

TH2

x+25=0

x=-25

Vậy x e { 38;-25}

Câu 2

= 4544 + 32 . (-7 - 13)

= 4544 + 32 . (-20)

= 4544 + (-640)

= 3904

@minhnguvn

Câu 1 :

\(\left(38-x\right).\left(x+25\right)=0\)

\(\orbr{\begin{cases}38-x=0\\x+25=0\end{cases}}\)

\(\orbr{\begin{cases}x=38\\x=-25\end{cases}}\)

Câu 2 :

\(71.64+32.\left(-7\right)-13.32\)

\(=4544+32.\left(-7\right)-13.32\)

\(=4544+32.\left(\left(-7\right)-13\right)\)

\(=4544+32.20\)

\(=4544+52\)

\(=4596\)