1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

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

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

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

\(=2x^3+6x\)

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

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

Bài 1 : 

a, \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)

\(=x^2+6x+9+x^2-6x+9+2x^2-18\)

\(=4x^2\)

b, \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9=8\)

Bài 2 : 

a, \(16x-8xy+xy^2=x\left(16-8y+y^2\right)=x\left(4-y\right)^2\)

b, \(3\left(3-x\right)-2x\left(x-3\right)=3\left(3-x\right)+2x\left(3-x\right)=\left(3+2x\right)\left(3-x\right)\)

c, \(3x^2+4x-4=3x^2+6x-2x-4=\left(x+2\right)\left(3x-2\right)\)

Rút gọn biểu thức

a) ( x + 3 )²  + ( x - 3 )² + 2( x² - 9 )

= x³ + 6x + 9 + x² - 6x + 9 + 2x² - 18

= 4x²

b) ( 4x - 1 )³ - ( 4x - 3 )( 16x² + 3 )

= 64x³ - 48x² + 12x - 1 - ( 64x³ - 48x² + 12x - 9 )

= 64x³ - 48x² + 12x - 1 - 64x³ + 48x² - 12x + 9

= 8

PTĐTTNT

a) 16x - 8xy + xy²

= x( 16 - 8y + y² )

= x( 4 - y )²

b) 3( 3 - x ) ± 2x( x - 3 ) < không biết thay dấu gì (: >

= 3( 3 - x ) \(\mp\)2x( 3 - x )

= ( 3 - x )( 3 \(\mp\)2x ) 

c) 3x² + 4x - 4

= 3x² + 6x - 2x - 4

= 3x( x + 2 ) - 2( x + 2 )

= ( x + 2 )( 3x - 2 )

Tìm x

a) ( 3x - 2 )( 3x + 4 ) - ( 2 - 3x )² = 6

<=> ( 3x - 2 )( 3x + 4 ) - ( 3x - 2 )² = 6

<=> ( 3x - 2 )( 3x + 4 - 3x + 2 ) = 6

<=> ( 3x - 2 ).6 = 6

<=> 3x - 2 = 1

<=> x = 1

b) 2( x - 3 ) - ( x - 3 )( 3x - 2 ) = 0

<=> ( x - 3 )( 2 - 3x + 2 ) = 0

<=> ( x - 3 )( 4 - 3x ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{3}\end{cases}}\)

c) ( x - 1 )( x + 2 ) - x( x - 2 ) = -5

<=> x² + x - 2 - x² + 2x = -5

<=> 3x - 2 = -5

<=> 3x = -3

<=> x = -1

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

\(\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=-4a^2\)

\(\left(a^2+b^2+c^2+a^2-b^2-c^2\right)\left(a^2+b^2+c^2-a^2+b^2+c^2\right)=2a^2\left(2b^2+2c^2\right)=4a^2b^2+4a^2c^2\)

\(\left(a-5\right)^2\left(a+5\right)^2=\left(a^2-25\right)^2\)

\(\left(3a^3+1\right)^2-9a^2-\left(3a^3+1\right)^2=-9a^2\)

Bài 8:

Ta có: \(A=-x^2+2x+4\)

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

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

\(=-\left(x-1\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=1