\(A=3x-x^2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{9}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

Vậy GTLN của A là \(\frac{9}{4}\)khi x = \(\frac{3}{2}\)

\(B=7-8x-x^2=-\left(x^2+8x+16\right)+23=-\left(x+4\right)^2+23\le23\)

Vậy GTLN của B là 23 khi x = -4

\(C=x^2-20x+101=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

Vậy GTNN của C là 1 khi x = 10

\(D=3x^2-6x+11=3\left(x^2-2x+1\right)+8=3\left(x-1\right)^2+8\ge8\)

Vậy GTNN của D là 8 khi x = 1

\(a,A=3x-x^2=-x^2+3x=-x^2+2.\frac{3}{2}x-\frac{9}{4}+\frac{9}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

Vậy Max A = 9/4 <=> x = 3/2

\(b,B=7-8x-x^2=-x^2-8x+7=-x^2-2.4x-16+23=-\left(x+4\right)^2+23\ge23\)

Vậy MinB = 23 <=> x = -4

\(c,C=x^2-20x+101=x^2-2.10x+10^2+1=\left(x-10\right)^2+1\ge1\)

Vậy MinC = 1 <=> x = 10

\(d,D=3x^2-6x+11\)

\(D=\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+\left(\sqrt{3}\right)^2+8=\left(\sqrt{3}x-\sqrt{3}\right)^2+8\ge8\)

Vậy MinD = 8<=> x=1

\(x^4-4x^3+4x^2\)

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

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

\(3x^2+10x+3\)

\(=3x^2+x+9x+3\)

\(=x\left(3x+1\right)+3\left(3x+1\right)\)

\(=\left(x+3\right)\left(3x+1\right)\)

\(x^4-4x^3+4x^2\)

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

\(=x^2.\left(x-2\right)^2\)

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

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

\(=3x^2y+3xy^2\)

\(=3xy\left(x+y\right)\)

\(a,\left(2x+3y\right)^2-4\left(2x+3y\right)\)

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

\(b,\left(x+y\right)^3-x^3-y^3\)

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

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

\(=\left(x+y\right).3x\)

\(c,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

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

\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)

a)  \(x\left(x-2\right)+x-2=0\)

<=>   \(\left(x-2\right)\left(x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy...

b)  \(5x\left(x-3\right)-x+3=0\)

<=>  \(\left(x-3\right)\left(5x-1\right)=0\)

<=>  \(\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)

Vậy...

\(a^5+b^5-\left(a+b\right)^5\)

\(=a^5+b^5-a^5-5a^4b-10a^3b^2-10a^2b^3-5ab^4-b^5\)

\(=-5ab\left(a^3+2a^2b+2ab^2+b^3\right)\)

\(=-5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)