Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

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

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

e) (x + 5)² - (2x - 1)²

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

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

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

(x - 10)(x + 2) = 0

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2

a) 3x . ( x-1 ) = 3x² - 3x 

b) x³- 2x²+x = x².( x-1 ) - x.( x-1 ) = (x-1).(x-1).x 

= (x-1)².x 

c) x²- 2xy-9z²+y²

= (x²-2xy+y² )-(3z)²

= (x-y)²-(3z)²

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

thay vào ta có ( 6+4-90 ).(6+4+90 )=-80.100=-8000 

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

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

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

a,

\(A=4(x-2)(x+1)+(2x-4)^2+(x+1)^2\\=[2(x-2)]^2+2\cdot2(x-2)(x+1)+(x+1)^2\\=[2(x-2)+(x+1)]^2\\=(2x-4+x+1)^2\\=(3x-3)^2\)

Thay $x=\dfrac12$ vào $A$, ta được:

\(A=\Bigg(3\cdot\dfrac12-3\Bigg)^2=\Bigg(\dfrac{-3}{2}\Bigg)^2=\dfrac94\)

Vậy $A=\dfrac94$ khi $x=\dfrac12$.

b,

\(B=x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\=(x^9-1)-(x^7-x^4)-(x^6-x^3)-(x^5-x^2)\\=[(x^3)^3-1]-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1)-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1-x^4-x^3-x^2)\\=(x^3-1)(x^6-x^4-x^2+1)\)

Thay $x=1$ vào $B$, ta được:

\(B=(1^3-1)(1^6-1^4-1^2+1)=0\)

Vậy $B=0$ khi $x=1$.

$Toru$

Phân tích đa thức thành nhân tử

a) 2( x + 1 ) - 3y( x + 1 ) = ( x + 1 )( 2 - 3y )

b) x² - 5x + 4 = x² - x - 4x + 4 = x( x - 1 ) - 4( x - 1 ) = ( x - 1 )( x - 4 )

Tìm x

a) x( x - 3 ) + 7x - 21 = 0

<=> x( x - 3 ) + 7( x - 3 ) = 0

<=> ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

b) ( x - 2 )² + x( 3 - x ) = 6

<=> x² - 4x + 4 + 3x - x² = 6

<=> -x + 4 = 6

<=> -x = 2

<=> x = -2

\(A=\frac{x-2}{x}\)và \(B=\frac{x}{x-2}-\frac{2x}{x^2-4}\)( x ≠ 0 ; x ≠ ±3 )

a) Tại x = 23 ( tmđk ) => \(A=\frac{23-2}{23}=\frac{21}{23}\)

b) P = A.B

\(=\frac{x-2}{x}\times\left(\frac{x}{x-2}-\frac{2x}{x^2-4}\right)\)

\(=\frac{x-2}{x}\times\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\frac{x-2}{x}\times\frac{x^2+2x-2x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{1}{x}\times\frac{x^2}{x+2}=\frac{x}{x+2}\)

Để P = 4 => \(\frac{x}{x+2}=4\)

=> 4( x + 2 ) = x

=> 4x + 8 - x = 0

=> 3x + 8 = 0

=> x = -8/3 ( tmđk )

Bài 1: 

a: \(49-4x^2=\left(7-2x\right)\left(7+2x\right)\)

b: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

c: \(x^2+18xy+81y^2=\left(x+9y\right)^2\)

Bài 3. a) x(x-2)-2x+x=0

       <=> x²-2x-2x+x=0

       <=>x²-4x+x=0

       <=>x²-3x=0

      <=> x(x-3)=0 => x=0; x=3.

 \(a,4x^2-1\)

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

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

\(b,25x^2-0,09\)

\(=\left(5x\right)^2-\left(0,3\right)^2\)

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

\(d,\left(x-y\right)^2-4\)

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

\(=\left(x-y-2\right)\left(x-y+2\right)\)

\(e,9-\left(x-y\right)^2\)

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

\(=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]\)

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

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

\(f,\left(x^2+4\right)^2-16x^2\)

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

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

\(=\left(x^2-2\cdot x\cdot2+2^2\right)\left(x^2+2\cdot x\cdot2+2^2\right)\)

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

#\(Toru\)

`#3107`

a)

`4x^2 - 1`

`= (2x)^2 - 1^2`

`= (2x - 1)(2x + 1)`

b)

`25x^2 - 0,09`

`= (5x)^2 - (0,3)^2`

`= (5x - 0,3)(5x + 0,3)`

d)

`(x - y)^2 - 4`

`= (x - y)^2 - 2^2`

`= (x - y - 2)(x - y + 2)`

e)

`9 - (x - y)^2`

`= 3^2 - (x - y)^2`

`= (3 - x + y)(3 + x - y)`

f)

`(x^2 + 4)^2 - 16x^2`

`= (x^2 + 4)^2 - (4x)^2`

`= (x^2 - 4x + 4)(x^2 + 4x + 4)`

`= (x - 2)^2 * (x + 2)^2`

_____

Tất cả các câu trên bạn sử dụng hđt:

`A^2 - B^2 = (A - B)(A + B)`

\(#MaiChangLaAnhDau..\)