Mấy bài này khá đơn giản .

Bạn chỉ cần áp dụng hằng đẳng thức \(x^2-y^2=\left(x+y\right)\left(x-y\right)\) là được nhs =))

a)

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

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

\(=\left[xy+4-2\left(x+y\right)\right]\left[xy+4+2\left(x+y\right)\right]\)

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

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

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

b)

\(\left(ab-xy\right)^2-\left(bx-ay\right)^2\)

\(=\left(ab-xy-bx+ay\right)\left(ab-xy+bx-ay\right)\)

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

\(=\left(b+y\right)\left(a-x\right)\left(a+x\right)\left(b-y\right)\)

c)

\(=\left(x^2+8x-34+3x^2-8x-2\right)\left(x^2+8x-34-3x^2+8x+2\right)\)

\(=\left(4x^2-36\right)\left(-2x^2+16x-32\right)\)

\(=\left(2x-6\right)\left(2x+6\right)\left(-2\right)\left(x^2-8x+16\right)\)

\(=\left(2x-6\right)\left(2x+6\right)\left(-2\right)\left(x-4\right)^2\)

Bạn liểm tra lại nhs

Mk lm hay nhấm lắm

=))

 ( x² + 8x - 34)² - ( 3x² - 8x + 2)²

=> x⁴ + 8x² - 34² - 3x⁴ + 8x² - 2²

=> x⁴( 1 - 4 ) + ( 8x² - 8x² ) + ( 34² - 2² )

-3x⁴ + 0 + 1152

= -3x⁴ + 1152

Lk t nhá, Michiel na`h, h pải dùng nik ko có tn

lẹ

gấp

Mày ra câu hỏi từ từ người ta trả lới cho chứ cứ hối người ta 😡

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

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

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

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

\(a,x^4+5x^3-8x-40=x^3\left(x+5\right)-8\left(x+5\right)\\ =\left(x^3-8\right)\left(x+5\right)=\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)\\ b,3x^2-6x-12y^2+3=3\left(x^2-2x-4y^2+1\right)\\ =3\left[\left(x-1\right)^2-4y^2\right]=3\left(x-2y-1\right)\left(x+2y-1\right)\)

\(x^2+8x-9\)

\(=x^2-x+9x-9\)

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

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

Câu đầu chưa học sorry

( x³ - 3x² + 3x -1 ) : ( x² - 2x + 1)

= ( x -1 )³ : ( x - 1 ) ²

= x -1

x² + 8x - 9

= x² -x + 9x - 9

= x ( x - 1 ) + 9 ( x- 1)

= ( x -1 ) ( x + 9)

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

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

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

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

x³ - 3x²y + 3xy² - y³ - z³

= (x³ - 3x²y + 3xy² - y³) - z³

= (x - y)³ - z³

= (x - y - z)[(x - y)² + (x - y)z + z²]

= (x - y - z)(x² - 2xy + y² + xz - yz + z³)

--------------------

x² - y² + 8x + 6y + 7

= (x² + 8x + 16) - (y² - 6y + 9)

= (x + 4)² - (y - 3)²

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

= (x - y + 7)(x + y + 1)

a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)

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

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

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

b: \(=x^2+8x+16-y^2+6y-9\)

=(x+4)^2-(y-3)^2

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

=(x+y+1)(x-y+7)