1) \(=\left(9x^2-25y^2\right)-\left(6x-10y\right)=\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)=\left(3x-5y\right)\left(3x+5y-2\right)\)

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

A) \(...=\left(7y-3\right)^3\)

B) \(...=\left(4y-3\right)^3\)

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

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

D) \(...=x^2-6x+9-\left(y^2-10y+25\right)\)

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

cậu có thể giải chi tiết giúp tớ dc ko

Áp dụng \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

              \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)

               \(\left(a+b\right)^2=a^2+2ab+b^2;\left(a-b\right)^2=a^2-2ab+b^2\)

Lần sau bạn chú ý viết đầy đủ yêu cầu của đề bài.

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

b) \(B=-x^2+4x+5=-\left(x^2-4x-5\right)=-\left(x-5\right)\left(x+1\right)\)

c) \(C=9x^2-6x+3=3\left(3x^2-2x+1\right)\)

Bài làm

a) xy + y² - x - y

= ( xy + y² ) - ( x + y )

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

= ( x + y )( y - 1 )

b) 25 - x² + 4xy - 4y²

= 25 - ( x² - 4xy + 4y² )

= 25 - ( x - 2y )²

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

c) xy + xz - 2y - 2z

= ( xy + xz ) - ( 2y + 2z )

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

= ( y + z )( x - 2 )

d) x² - 6xy + 9y² - 25z²

= ( x² - 6xy + 9y² ) - 25z²

= ( x - 3y )² - 25z²

= ( x - 3y - 5z )( z - 3y + 5z )

e) 3x² - 3y² - 12x + 12y

= 3( x - y )( x + y ) - 12( x - y )

= ( x - y )[ 3( x + y ) - 12 ]

f) 4x³ + 4xy² + 8x²y - 16x

= 4x( x² + y² + 2xy - 4 )

= 4x[ ( x + y)² - 4 ]

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

g) x² - 5x + 4

= x² - x - 4x + 4

= x( x - 1 ) - 4( x - 1 )

= ( x - 1 )( x - 4 )

h) x⁴ + 5x² + 4

= x⁴ + x² + 4x² + 4

= x²( x² + 1 ) + 4( x² + 1 )

= ( x² + 1 )( x² + 4 )

i) 2x² + 3x - 5

= 2x² - 5x + 2x - 5

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

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

k) x³ - 2x² + 6x - 5 ( không biết làm )
l) x² - 4x + 3

= ( x² - 4x + 4 ) - 1

= ( x - 2 )² - 1

= ( x - 3 )( x - 1 )

# Học tốt #

Bài 4: 

a: Xét tứ giác ABDC có 

M là trung điểm của BC

M là trung điểm của AD

Do đó: ABDC là hình bình hành

Suy ra: BD=AC

b: Ta có: ABDC là hình bình hành

nên AB//DC

a) \(x^3+x^2y-x^2z-xyz\)

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

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

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

b) \(x^2-6x+9-9y^2\)

\(=\left(x^2-2\cdot x\cdot3+3^2\right)-\left(3y\right)^2\)

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

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

c) \(x^2+9x+20\)

\(=x^2+5x+4x+20\)

\(=x\left(x+5\right)+4\left(x+5\right)\)

\(=\left(x+5\right)\left(x+4\right)\)

d) \(x^4+4\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot2+4-2\cdot x^2\cdot2\)

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

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

a/\(x^3+x^2y-x^2z-xyz\)

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

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

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

b/\(x^2-6x+9-9y^2\)

\(=\left(x^2-6x+9\right)-9y^2\)

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

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

c/\(x^2+9x+20\)

\(=x^2+4x+5x+20\)

\(=\left(x^2+4x\right)+\left(5x+20\right)\)

\(=x\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+5\right)\left(x+4\right)\)

d/\(x^4+4\)

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

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

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

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

\(a,x^3+x^2y-x^2z-xyz=x^2\left(x+y\right)-xz\left(x+y\right).\)

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

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

\(b,x^2-6x+9-9y^2=\left(x-3\right)^2-9y^2\)

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

\(c,x^2+9x+20=\left(x^2+8x+16\right)+\left(x+4\right)\)

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

\(=\left(x+4\right)\left(x+4+1\right)=\left(x+4\right)\left(x+5\right)\)

\(d,x^4+4=\left(x^2\right)^2+2^2\)

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

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