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

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1

\(y=\frac{\frac{^x}{x^2}-x-6-x-\frac{1}{3}x^2-4x-15}{x^4}-2x^2+\frac{1}{3}x^2+11x+10b\)

\(y=\frac{-\left(5x^7-33x^6-30bx^5+x^3+18x^2+63x-3\right)}{3x^5}\)

+) \(P=3x-6-3x+5=-1\)

+) \(Q=2x-8+3x+8=5x\)

+) R bạn xem lại điều kiện

Ta có: P=|3x-6|-3x+5

=3x-6-3x+5

=-1

Ta có: Q=|8-2x|+3x+8

=2x-8+3x+8

=5x

(3x+4)² +(4x -1 )² + 2x + 5(2x-5)

= 9x²+24x+16 + 16x²-8x+1 + 2x + 10x - 25

= 25x² + 28x - 8

b. (2x+1)(4x² -2x + 1 ) + ( 2-3x ) ( 4+ 6x + 9x² )

= 8x³ + 1 + 8-27x³

= -19x³ +9

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

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