lấy 4xy làm nhân tử chung rồi tính nha

hk tốt nha bn

1.

\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

\(=5x^2-3xy^2+4y\)

2.

a)  \(27x^4-8x=x\left(27x^3-8\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

c) \(x^2-2x-5+2\sqrt{5}\)

\(=\left(x-1\right)^2-6+2\sqrt{5}\)

\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)

Bài 1:

 \(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)

\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

\(=5x^2-3xy^2+4y\)

Bài 2: 

a) \(27x^4-8x\)

\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b) \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)

\(=3xy\left(x-y\right)-12\left(x-y\right)\)

\(=\left(3xy-12\right)\left(x-y\right)\)

2) \(4x^3+4xy^2+8x^2y-16x\)

\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)

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

Ta có : 3x² - 3y² - 12x + 12y 

= (3x² - 3y²) - (12x - 12y)

= 3(x² - y²) - 12(x - y)

= 3(x - y)(x + y) - 4.3.(x - y)

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

3) x^4-5x^2+4 = x^4-x^2-4x^2+4 

                        =x^2(x^2-1)-4(x^2-1)

                        =(x^2-1)(x^2-4)

đa thức lớp 5 hả bạm

đa thức lớp 5 à

mình ghi sao đề, các bạn ko cần làm đâu

Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$

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$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$

$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.

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

$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$

$=3(x-2)(x^2+2x+4)-240x$

$=3(x^3-2^3)-240x=3x^3-240x-24$

$=3(x^3-80x-8)$

\(5\left(x^2+y^2\right)^2-20x^2y^2\)

\(=5x^4+5y^4-10x^2y^2\)

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

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

Lời giải:

$5(x^2+-y^2)^2-20x^2y^2=5(x^2-y^2)^2-20x^2y^2$

$=5[(x^2-y^2)^2-(2xy)^2]=5(x^2-y^2-2xy)(x^2-y^2+2xy)$

mik ko bít

I don't now

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4x²+4y-4xy-3y²-1

=(4x²-4xy+y²)-(4y²-4y+1)

=(2x-y)²-(2y-1)²

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

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

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

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

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

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