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

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

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

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

Lời giải:

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

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

\(x^4+x^3-20x^2-47x-15\)

\(=x^3\left(x-5\right)+6x^2\left(x-5\right)+10x\left(x-5\right)+3\left(x-5\right)\)

\(=\left(x-5\right)\left(x^3+6x^2+10x+3\right)\)

\(=\left(x-5\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)+\left(x+3\right)\right]\)

\(=\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)

\(=x^4-5x^3+6x^3-30x^2+10x^2-50x+3x-15\\ =\left(x-5\right)\left(x^3+6x^2+10x+3\right)\\ =\left(x-5\right)\left(x^3+3x^2+3x^2+9x+x+3\right)\\ =\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)

x⁶y - 5x⁵ - 4x⁴y + 20x³ 

= ( x⁶y - 5x⁵ ) - ( 4x⁴y - 20x³ ) 

= x⁵( xy - 5 ) - 4x³( xy - 5 )

= ( x⁵ - 4x³ )( xy - 5 ) = x³( x² - 4 )( xy - 5 ) 

= x³( x - 2 )(x + 2 )( xy - 5 )

 = x^3.(x^3y-5x^2-4xy+20)

 = x^3.[(x^3y-5x^2)-(4xy-20)]

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

k mk nha

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

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

=>x=-3 hoặc x=3

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.

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

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

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

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

\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2-a^2xy-b^2xy\)

\(=\left(abx^2-b^2xy\right)-\left(a^2xy-aby^2\right)\)

\(=bx\left(ax-by\right)-ay\left(ax-by\right)\)

\(=\left(ax-by\right)\left(bx-ay\right)\)

5²y² + 10xy³ - 20xy²

= 5y.(5y + 2xy² - 4xy)