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

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

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

= ( x³ + 3x²y + 3xy² + y³ ) - 6xy - 3x² - 3y² + 3x + 3y + 2012

= ( x + y )³ - 3xy - 3x² - 3xy - y² + 3. ( x + y ) + 2012

= ( x + y )³ - 3x ( x + y ) - 3y .( x + y ) + 3.( x + y ) + 2012

= ( x + y )³ - 3.( x + y ) ( x + y ) + 3( x + y ) + 2012

= 101³ - 3.101² + 3.101 + 2012

= 1002013

\(5x-\frac{3-2x}{2}>\frac{7x-5}{2}+x\)

\(\Leftrightarrow\) \(\frac{10x}{2}-\frac{3-2x}{2}>\frac{7x-5}{2}+\frac{2x}{2}\)

\(\Rightarrow\) \(10x-3+2x>7x-5+2x\)

\(\Leftrightarrow\) \(10x+2x-7x-2x>-5+3\)

\(\Leftrightarrow\) \(3x>-2\) 

\(\Leftrightarrow\) \(x>-\frac{2}{3}\)

Vậy ................

4x > (5x -2)/2

4x - 5x/2 > -1

3x/2 > -1

3x > -2

x>-2/3

1, a ( a - b ) ( a + b )  - ( a + b ) ( a² - ab + b² )

= ( a + b ) [ a ( a - b ) - ( a² - ab + b² )

= ( a + b ) ( a² - ab - a² + ab - b² )

= ( a + b ) b²

.......

2, 3x ( x + 7 )² - 11x² ( x + 7 ) + 9 ( x + 7 )

= ( x + 7 ) [( 3x ( x + 7 ) - 11x² + 9 ]

= ( x + 7 ) ( 3x³ + 21x - 11x² + 9)

= ( x + 7 ) ( - 8x² + 21x + 9 )

..........

3, 4x ( x - 2y ) + 8y ( 2y - x )

= 4x ( x - 2y ) - 8y ( x - 2y )

=  ( 4x - 8y ) ( x - 2y )

= 4 ( x - 2y ) ( x - 2y )

= 4 ( x - 2y )²

1,a(a-b)(a+b)-(a+b)(a²-ab+b²)=(a+b)[a(a-b)-a²+ab-b²]

                                                =(a+b)(a²-ab-a²+ab-b²)

                                                =(a+b)(-b²)=-b²(a+b)

\(p^{m+2}.q-p^{m+1}.q^3-p^2.q^{n+1}+p.q^{n+3}\)

\(=pq\left(p^{m+1}-p^mq^2-pq^n+q^{n+2}\right)\)

\(=p^m\left(p-q^2\right)-q^n\left(p-q^2\right)\)

\(=\left(p-q^2\right)\left(p^m-q^n\right)\)

..........

\(a^3+b^3+c^3=3abc\) 

<=>   \(a^3+b^3+c^3-3abc=0\)

<=>    \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>    \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

  Xét:     \(a^2+b^2+c^2-ab-bc-ca=0\)

<=>    \(2a^{ 2}+2b^2+2c^2-2ab-2bc-2ca=0\)

<=>     \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>    \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\) <=>  \(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)<=>   \(a=b=c\)

=>  đpcm