\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)

\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)

A

D bạn nhé

a) \(5x^3-10x^2+15x=5x\left(x^2-2x+3\right)\)

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

a: \(=5x\left(x^2-2x+3\right)\)

b: =(x-1)(x-2)

a) 5 𝑥 3 − 10 𝑥 2 + 15 𝑥 5x 3 −10x 2 +15x Bước 1: Tìm ước chung lớn nhất của các hệ số và biến: Hệ số: 5, 10, 15 → ước chung lớn nhất là 5 Biến: 𝑥 3 , 𝑥 2 , 𝑥 x 3 ,x 2 ,x → lấy 𝑥 x (lũy thừa nhỏ nhất) Bước 2: Lấy 5 𝑥 5x làm nhân tử chung: 5 𝑥 3 − 10 𝑥 2 + 15 𝑥 = 5 𝑥 ( 𝑥 2 − 2 𝑥 + 3 ) 5x 3 −10x 2 +15x=5x(x 2 −2x+3) b) 𝑥 2 − 3 𝑥 + 2 x 2 −3x+2 Ta tìm hai số 𝑎 , 𝑏 a,b sao cho: 𝑎 + 𝑏 = − 3 a+b=−3 𝑎 × 𝑏 = 2 a×b=2 Hai số đó là − 1 −1 và − 2 −2 (vì − 1 + − 2 = − 3 −1+−2=−3, − 1 × − 2 = 2 −1×−2=2). Vậy: 𝑥 2 − 3 𝑥 + 2 = ( 𝑥 − 1 ) ( 𝑥 − 2 ) x 2 −3x+2=(x−1)(x−2) Kết luận: { 5 𝑥 3 − 10 𝑥 2 + 15 𝑥 = 5 𝑥 ( 𝑥 2 − 2 𝑥 + 3 ) 𝑥 2 − 3 𝑥 + 2 = ( 𝑥 − 1 ) ( 𝑥 − 2 ) { 5x 3 −10x 2 +15x=5x(x 2 −2x+3) x 2 −3x+2=(x−1)(x−2) ​ ​ khó đấy

a: \(x^4+2x^3+3x^2+2x+1\)

\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)

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

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

b: \(x^4-4x^3+2x^2+4x+1\)

\(=x^4-2x^3-x^2-2x^3+4x^2+2x-x^2+2x+1\)

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

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

c: \(x^4+x^3+2x^2+2x+4\)

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

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

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

1+1 = 1+1

Lời giải:
a.

$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)[(x^2+1)^2-x^2]$

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

b. 

$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$

$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$

c.

$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$

$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$

$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$

c.

$x^2-5y^2-y^4+2xy-9$

$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$

\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)

a) x⁸+x⁴+1 = (x⁴+1)²-x⁴ = (x⁴-x²+1)(x⁴+x²+1)

b) x¹²-3x⁶-1 = (x⁶-1)²-x⁶ = (x⁶-x³-1)(x⁶+x³-1)

c) 3x⁴+10x²-25 = 4x⁴-(x⁴-10x²+25) = 4x⁴- (x²-5)² = (x²+5)(3x²-5)

d) x²-5y²-y⁴+2xy-9 = (x+y)²-(y²+3)² = (x+y-y²-3)(x+y+y²+3)

a ) x=0; x = -(căn bậc hai(7)*i-3)/8;x = (căn bậc hai(7)*i+3)/8;

b ) -(y-x-3)*(y+x+3)

a) \(12x^3-9x^2+3x\)

\(=3x\left(4x^2-3x+1\right)\)

b) \(x^2-y^2+6x+9\)

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

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

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

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

2, \(5x\left(3y+4x-6\right)\)

3, \(3x\left(2-y\right)\)

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

5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)

6, \(2xy\left(x+2y-5x^2y\right)\)

7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

11, \(\left(x+y\right)\left(x-1\right)\)

\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)

a:

=x(x²-y²-10x+25)

=x((x²-10x+25)-y²)

=x((x-5)²-y²)

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

b

=>8x(x-5)-3(x-5)=0

=>(x-5)(8x-3)=0

x-5=0=>x=5 hoặc 8x-3=0=>x=3/8