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Những câu hỏi liên quan
ngọc hân
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Nguyễn Lê Phước Thịnh
26 tháng 8 2021 lúc 22:28

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

b: \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

c: \(x^8+x^4+1\)

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

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

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

 

linh phạm
26 tháng 8 2021 lúc 20:30

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

 

DarkWitch
26 tháng 8 2021 lúc 20:31

\(a)\; x^4+4 \\= x^4+4x^2+4-4x^2\\=(x^2+2)^2-4x^2\\=(x^2+2-2x)(x^2+2+2x)\)

nè Moon
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Yeutoanhoc
1 tháng 10 2021 lúc 9:45

`a)x^3-8x^2+16x`

`=x(x^2-8x+16)`

`=x(x-4)^2`

`b)x^2+4y^2+2x-4y-4xy-24`

`=(x-2y)^2+2(x-2y)-24`

`=(x-2y)^2-4(x-2y)+6(x-2y)-24`

`=(x-2y-4)(x-2y+6)`

`c)x^4+x^3-x^2-2x-2`

`=x^4-2x^2+x^3-2x+x^2-2`

`=x^2(x^2-2)+x(x^2-2)+x^2-2`

`=(x^2-2)(x^2+x+1)`

ThanhNghiem
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๖ۣۜHả๖ۣۜI
4 tháng 8 2023 lúc 14:25

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

\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)

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

\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)

 

 

 

Nguyễn Lê Phước Thịnh
4 tháng 8 2023 lúc 14:15

b: =xy-x-y+1

=x(y-1)-(y-1)

=(x-1)(y-1)

c: =(x-2y)^2-4y

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

d: =16-(x^2-2xy+y^2)

=16-(x-y)^2

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

mit béo
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ILoveMath
19 tháng 12 2021 lúc 10:03

\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)

\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)

Nguyễn Thảo Nguyên
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Nguyễn Hoàng Minh
12 tháng 7 2023 lúc 21:56

\(a,=\left(5x^3+10x\right)+\left(x^4-4\right)\\ =5x\left(x^2+2\right)+\left(x^2+2\right)\left(x^2-2\right)\\ =\left(x^2+2\right)\left(x^2+5x-2\right)\\ b,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(c,=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\\ d,=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\\ e,=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^{10}-x^7+x^5-x^4+x^3-x+1\right)\)

Nguyễn Lê Phước Thịnh
12 tháng 7 2023 lúc 21:54

a: =x^4+2x^2+5x^3+10x-2x^2-4

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

b; =(x+y)^3+z^3-3xy(x+y)-3xyz

=(x+y+z)*(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)

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

c: =x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1

=(x^2+x+1)(x^6-x^5+x^3-x^2+1)

Daisy
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ILoveMath
13 tháng 11 2021 lúc 20:43

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

b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)

c) đặt \(x^2+x+1=t\)

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

Trần Hương Trà
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ILoveMath
23 tháng 8 2021 lúc 9:35

\(A=x^2+4=\left(x^2+4x+4\right)-4x=\left(x+2\right)^2-\sqrt{4x}=\left(x+2-\sqrt{4x}\right)\left(x+2+\sqrt{4x}\right)\)

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

Trên con đường thành côn...
23 tháng 8 2021 lúc 9:36

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Lê Đăng Hải Phong
Xem chi tiết
Nguyễn Lê Phước Thịnh
7 tháng 11 2021 lúc 21:47

a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)

b: \(=25-\left(x-2y\right)^2\)

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

Đinh Cẩm Tú
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Lấp La Lấp Lánh
7 tháng 9 2021 lúc 9:51

a) \(x^2+5x+4==x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

b) \(3x^2+4x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(3x+7\right)\)

c) \(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

Rin Huỳnh
7 tháng 9 2021 lúc 9:51

a) (x+1)(x+4)

b) (x-1)(3x+7)

c) (x+3)(x+4)

Edogawa Conan
7 tháng 9 2021 lúc 9:52

a) x2+5x+4 = x(x+4)+(x+4) = (x+4)(x+1)

b) 3x2+4x-7 = 3x(x-1)+7(x-1) = (x-1)(3x+7)

c) x2+7x+12 = x(x+4)+3(x+4) = (x+3)(x+4)

Mitt Chuche
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Kiều Vũ Linh
17 tháng 12 2023 lúc 16:53

a) x³y + x - y - 1

= (x³y - y) + (x - 1)

= y(x³ - 1) + (x - 1)

= y(x - 1)(x² + x + 1) + (x - 1)

= (x - 1)[y(x² + x + 1) + 1]

= (x - 1)(x²y + xy + y + 1)

b) x²(x - 2) + 4(2 - x)

= x²(x - 2) - 4(x - 2)

= (x - 2)(x² - 4)

= (x - 2)(x - 2)(x + 2)

= (x - 2)²(x + 2)

c) x³ - x² - 20x

= x(x² - x - 20)

= x(x² + 4x - 5x - 20)

= x[(x² + 4x) - (5x + 20)]

= x[x(x + 4) - 5(x + 4)]

= x(x + 4)(x - 5)

d) (x² + 1)² - (x + 1)²

= (x² + 1 - x - 1)(x² + 1 + x + 1)

= (x² - x)(x² + x + 2)

= x(x - 1)(x² + x + 2)

Kiều Vũ Linh
17 tháng 12 2023 lúc 17:09

e) 6x² - 7x + 2

= 6x² - 3x - 4x + 2

= (6x² - 3x) - (4x - 2)

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

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

f) x⁴ + 8x² + 12

= x⁴ + 2x² + 6x² + 12

= (x⁴ + 2x²) + (6x² + 12)

= x²(x² + 2) + 6(x² + 2)

= (x² + 2)(x² + 6)

g) (x³ + x + 1)(x³ + x) - 2

Đặt u = x³ + x

x³ + x + 1 = u + 1

(u + 1).u - 2

= u² + u - 2

= u² - u + 2u - 2

= (u² - u) + (2u - 2)

= u(u - 1) + 2(u - 1)

= (u - 1)(u + 2)

= (x³ + x - 1)(x³ + x + 2)

= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)

= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]

= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]

= (x³ + x - 1)(x - 1)(x² - x + 2)

h) (x + 1)(x + 2)(x + 3)(x + 4) - 1

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1

= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)

Đặt u = x² + 5x + 4

u + 2 = x² + 5x + 6

(1) u.(u + 2) - 1

= u² + 2u - 1

= u² + 2u + 1 - 2

= (u² + 2u + 1) - 2

= (u + 1)² - 2

= (u + 1 + √2)(u + 1 - √2)

= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)

= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)

Nguyễn Lê Phước Thịnh
17 tháng 12 2023 lúc 21:05

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

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

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

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

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

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

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

l: \(81x^4+4y^4\)

\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)

\(=\left(81x^4+36x^2y^2+4y^4\right)-\left(6xy\right)^2\)

\(=\left[\left(9x^2\right)^2+2\cdot9x^2\cdot2y^2+\left(2y^2\right)^2\right]-\left(6xy\right)^2\)

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

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