Mình sửa: Bài 1
2)x²+3x-15

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

\(x^2-7x+10\)

\(=x^2-2x-5x+10\)

\(=x\left(x-2\right)-5\left(x-2\right)\)

\(=\left(x-2\right)\left(x-5\right)\)

học tốt

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

= x(x - 1 - y)(x - 1 + y)

= x(x - y - 1)(x + y - 1)

e) 27 - 3x²

= 3(9 - x²)

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

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

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

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

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

=>\(x\left(x+2\right)-3\left(x+2\right)=0\)

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

=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

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

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

\(=x\left[\left(x-1\right)^2-y^2\right]\)

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

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) 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 = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

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)

Bài 1 :

Tự phân tích vế trái và điền vào vế phải 

Bài 2 :

a) \(3x^3-6x^2+3x\)

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

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

b) \(2xy+z+2x+yz\)

\(=\left(2xy+2x\right)+\left(z+yz\right)\)

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

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

c) \(x^4-y^4\)

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

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

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

d) \(3x^2-4x-7\)

\(=3x^2+3x-7x-7\)

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

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

1

a, 2x²+4x+2-2y² = 2(x²+2x+1-y²)= 2[(x+1)²-y² ] = 2(x-y+1)(x+y+1)

b, 2x - 2y - x² + 2xy - y²= 2(x -y) - (x² - 2xy + y²) = 2(x-y)-(x-y)²=(x-y)(2-x+y)

c, x²-y²-2y-1=x²-(y²+2y+1)=x²-(y+1)²=(x-y-1)(x+y+1)

d, x²-4x-2xy-4y+y²= x²-2xy+y²-4x-4y=(x-y)

2.

a, x²-3x+2=x²-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)

b, x²+5x+6=x²+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)

c, x²+6x-6=

1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)

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

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

3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)

4, sửa đề : 

 \(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

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

a) x²(x-3)-4x+12

=x²(x-3)-4(x-3)

=(x-3)(x²-4)

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

b) 2a(x+y)-x-y

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

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

c) 2x-4+5x²-10x

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

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

d) 5x²-12x-7x+14

=5x²-19x+14

e) xy-y²-3x+3y

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

=(x-y)(y-3)

#H

Trả lời:

1, x² ( x - 3 ) - 4x + 12 = x² ( x - 3 ) - 4 ( x - 3 ) = ( x - 3 )( x² - 4 ) = ( x - 3 )( x - 2 )( x + 2 )

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

3, 2x - 4 + 5x² - 10x = ( 2x - 4 ) + ( 5x² - 10x ) = 2 ( x - 2 ) + 5x ( x - 2 ) = ( x - 2 )( 2 + 5x )

4, Sửa đề: 6x² - 12x - 7x + 14 = ( 6x² - 12x ) - ( 7x - 14 ) = 6x ( x - 2 ) - 7 ( x - 2 ) = ( x - 2 )( 6x - 7 )

5, xy - y² - 3x + 3y = ( xy - y² ) - ( 3x - 3y ) = y ( x - y ) - 3 ( x - y ) = ( x + y )( y - 3 )

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

=2x(x-3)+5(x-3)

=(x-3)(2x+5)

c: \(x^3-x^2+x-1\)

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

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

d: \(2x^3-5x-6\)

\(=2x^3-4x^2+4x^2-8x+3x-6\)

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

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

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

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

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

f; \(x^7+x^5+x^3\)

\(=x^3\left(x^4+x^2+1\right)\)

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

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

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

g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

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

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

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

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

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

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

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

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

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

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

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

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

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

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

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

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