Sorry , sửa lại :

\(\le\) \(\Rightarrow\) \(\ge\)

\(A=9x^2+18x-20\)

\(\Leftrightarrow A=\left(3x\right)^2+2.3x.3+9-29\)

\(\Leftrightarrow A=\left(3x+3\right)^2-29\le-29\forall x\)

Dấu " = " xảy ra

\(\Leftrightarrow\left(3x+3\right)^2=0\Leftrightarrow3x+3=0\Leftrightarrow3x=-3\Leftrightarrow x=-1\)

Vậy Min A là : \(-29\Leftrightarrow x=-1\)

\(B=m^2+10m+1\)

\(\Leftrightarrow B=m^2+2.m.5+25-24\)

\(\Leftrightarrow B=\left(m+5\right)^2-24\le-24\forall m\)

Dấu \("="\) xảy ra

\(\Leftrightarrow\left(m+5\right)^2=0\Leftrightarrow m+5=0\Leftrightarrow m=-5\)

Vậy Min B là : -24 \(\Leftrightarrow m=-5\)

\(C=25x^2-20x+30\)

\(\Leftrightarrow C=\left(5x\right)^2-2.5x.2+4+26\)

\(\Leftrightarrow C=\left(5x-2\right)^2+26\le26\forall x\)

Dấu " = " xảy ra

\(\Leftrightarrow\left(5x-2\right)^2=0\Leftrightarrow5x-2=0\Leftrightarrow5x=2\Leftrightarrow x=\dfrac{2}{5}\)

Vậy Min C là : 26 \(\Leftrightarrow x=\dfrac{2}{5}\)

P = ...

\(\Leftrightarrow P=x^3z-x^3y^2+y^3x-y^3z^2+z^3y-z^3x^2+x^2y^2z^2-xyz\)\(\Leftrightarrow P=\left(x^3z-x^2z^3\right)-\left(x^3y^2-x^2y^2z^2\right)+\left(xy^3-y^3z\right)+\left(yz^3-xyz\right)\)\(\Leftrightarrow P=x^2z\left(x-z^2\right)-x^2y^2\left(x-z^2\right)+y^3\left(x-z^2\right)-yz\left(x-z^2\right)\)\(\Leftrightarrow P=\left(x-z^2\right)\left(x^2z-x^2y^2+y^3-yz\right)\)

\(\Leftrightarrow P=\left(x-z^2\right)\left[\left(x^2z-x^2y^2\right)+\left(y^3-yz\right)\right]\)

\(\Leftrightarrow P=\left(x-z^2\right)\left[-x^2\left(y^2-z\right)+y\left(y^2-z\right)\right]\)

\(\Leftrightarrow P=\left(x-z\right)^2\left(y^2-z\right)\left(y-x^2\right)\)

\(\Leftrightarrow P=abc\left(đpcm\right)\)

f ) \(4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy ...

h ) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy ...

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-2-x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

Vậy ...

K ) \(\left(1-x\right)^2-1+x=0\)

\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0\)

\(\Leftrightarrow\left(1-x\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

Vậy ...

i ) \(\left(x-3\right)^3+3-x=0\)

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x-3=1\\x-3=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

Vậy ...

m ) \(x+6x^2=0\)

\(\Leftrightarrow x\left(1+6x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{6}\end{matrix}\right.\)

Vậy ...

n ) \(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy ...

a ) \(5\left(x+3\right)-6x-2x^2=0\)

\(\Leftrightarrow5\left(x+3\right)-2x\left(3+x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy ...

b ) \(\left(x-2004\right)=8016x-4x^2\)

\(\Leftrightarrow x-2004=-4x\left(x-2004\right)\)

\(\Leftrightarrow x-2004+4x\left(x-2004\right)=0\)

\(\Leftrightarrow\left(x-2004\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2004=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2004\\4x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2004\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy ...

c ) \(\left(x+1\right)^2=x+1\)

\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy ...

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

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

\(\dfrac{1}{5}m^2n^3:\left(-5m^2n^2\right)=\left(\dfrac{1}{5}:-5\right)\left(m^2:m^2\right)\left(n^3:n^2\right)=-\dfrac{1}{25}n\)

\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)

\(=\left(x^2-x+1\right)^2-2\left(x^2-x+1\right)\dfrac{5}{2}x+\left(\dfrac{5}{2}x\right)^2-\left(\dfrac{3}{2}x\right)^2\)

\(=\left(x^2-x+1-\dfrac{5}{2}x\right)^2-\left(\dfrac{3}{2}x\right)^2\)

\(=\left(x^2-\dfrac{7}{2}x+1\right)^2-\left(\dfrac{3}{2}x\right)^2\)

\(=\left(x^2-\dfrac{7}{2}x+1-\dfrac{3}{2}x\right)\left(x^2-\dfrac{7}{2}x+1+\dfrac{3}{2}x\right)\)

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

a ) \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)

\(=\left(x^2+x\right)^2+2\left(x^2+x\right)\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2\)

\(=\left(x^2+x+\dfrac{3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2\)

\(=\left(x^2+x+\dfrac{3}{2}-\dfrac{1}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{1}{2}\right)\)

\(=\left(x^2+x+1\right)\left(x^2+x+2\right)\)

b ) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)

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

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

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

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

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

\(x^2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

\(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)