dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

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

  \(=12x^2-8x-12x^2-3x\)

  \(=-11x\)       \(\left(1\right)\)

     Thay \(x=-2\) vào  \(\left(1\right)\) ta được :

            \(-11.\left(-2\right)=22\)

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

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

   \(=x^2-9-x^2+2x-1\)

   \(=2x-10\)       \(\left(2\right)\)

     Thay \(x=6\) vào \(\left(2\right)\) ta được :

             \(2.6-10=2\)

Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

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

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

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

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

g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

B1 :

a) (2x - 1)²