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 \(A_{min}=-34\Leftrightarrow x=5\)

Bài 8:

a) Ta có: \(A=\left(x-y\right)^3+3xy\left(x-y\right)\)

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

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

\(=x^3-y^3\)

b) Ta có: \(B=\left(x+y\right)^3+3\left(x-y\right)\left(x+y\right)^2+3\left(x-y\right)^2\left(x+y\right)+\left(x-y\right)^3\)

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

\(=\left(2x\right)^3=8x^3\)

3: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)=27\)

\(\Leftrightarrow x^3-27-x^3+x=27\)

hay x=54

c: \(C=\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

=4xy

d: \(D=\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)=\dfrac{1}{4}x^2-y^2\)