`a)7x^3y^2+14x^2y^3+7xy^4`

`=7xy^2(x^2+2xy+y^2)`

`=7xy^2(x+y)^2`

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`b)x^2-xy+5x-5y`

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

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

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`c)3x^2-6xy-12+3y^2`

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

`=3[(x-y)^2-4]`

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

a)7x³y²+14x²y³+7xy⁴

=7xy²(x²+2xy+y²)

=7xy²(x+y)²

b)x²-xy + 5x - 5y

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

=(x-y) (x+5)

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a) \(39x-39y=39\left(x-y\right)\)

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

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

c) \(16x^2+24xy+9y^2=\left(4x\right)^2+4x.3y.2+\left(3y\right)^2=\left(4x+3y\right)^2\)

d) \(25x^2-\frac{1}{25y^2}=\left(5x\right)^2-\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)\left(5x+\frac{1}{5y}\right)\)

e) \(7x^2-7xy+5x-5y=7x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(7x+5\right)\)

f) \(5x^2-45y^2-30y-5=5\left(x^2-9y^2-6y-1\right)=5\left[x^2-\left(9y^2+6y+1\right)\right]\)

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

g) \(x^2+2x+1-y^2-4y-1=\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\) ( Chắc đề vậy :v ) 

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

h) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)

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

Vô đây xem: bài 1:phân tích đa thức thành nhân tửa)7x^3y-14x^2y+7xy^3b)3x^2-3xy-5x+5yc)x^2+7x+12giúp mình với  - Hoc24

\(a,=7xy\left(x^2-2xy+y^2\right)=7xy\left(x-y\right)^2\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\)

a)  \(\left(x+y\right)^3-1\)

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

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

b)  \(x^2+2xy+x+2y\)

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

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

a/  (5x^2 -10xy+5y^2)-20z^2=5[(x-y)^2-(2z)^2]=5(x-y-2x)(x-y+2z)

b/16x-5x^2-3=-[5x^2-16x+3]=-[(5x^2-x)-(15x+3)]=-[x(5x-1)-3(5x-1)]=(3-x)(5x-1)

c/x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)

2a/ 5x(X^2-9)=0=>x=0 hoặc x^2=9=>x=0 hoặc x=+-3

b/x^2-7x+10=0=>(x^2-2x)-(5x-10)=0=>x(x-2)-5(x-2)=0=>x-2=0 hoặc x-5 =0 => tự tính nhé!

Answer:

Bài 1:

\(5x² - 10xy + 5y² - 20z²\)

\(= 5( x² - 2xy + y² - 4z²)\)

\(= 5 [(x² - 2xy + y²) - (2z)²]\)

\(= 5 [(x - y)² - (2z)²]\)

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

\(16x - 5x² - 3 \)

\(= -( 5x² - 16x + 3)\)

\(= -( 5x² - 15x - 1x + 3)\)

\(= - [ (5x² -x) - (15x -3)]\)

\(= - [ x(5x -1) -3(5x -1)]\)

\(= - (5x-1)(x-3)\)

\(x² + 4x + 3\)

\(= x² + x + 3x + 3\)

\(= (x² + x) + (3x + 3)\)

\(= x( x + 1) +3 (x+1)\)

\(= (x+1) (x+3)\)

Bài 2:

\(5x\left(x^2-9\right)=0\)

\(\Rightarrow5x\left(x-3\right)\left(x+3\right)=0\)

Trường hợp 1: \(5x=0\Leftrightarrow x=0\)

Trường hợp 2: \(x-3=0\Leftrightarrow x=3\)

Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)

\(x^2-7x+10=0\)

\(\Rightarrow x^2-5x-2x+10=0\)

\(\Rightarrow x\left(x-5\right)-2\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)

a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)

b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)

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

d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(x-2)

a) = (x - 1)(5x - 3)

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

= (x + y)(x - 5)

c) = x(x - y) - y(x - y)

= (x - y)^2

d) = 2(x² + 2x + 1 - y²)

= 2[(x + 1)² - y²]

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