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Phương Nhung Hà
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cau be ngoc
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Hermione Granger
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Nguyễn Lê Phước Thịnh
29 tháng 11 2023 lúc 21:57

a: \(x^4-2x^3+x^2-2x\)

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

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

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

b: \(x^4+x^3-8x-8\)

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

\(=x^3\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3-8\right)\)

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

Đặng Đình Tiến
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Toru
25 tháng 8 2023 lúc 20:51

\(b,x^3-2x^2-4xy^2+x\)

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

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

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

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

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

\(---\)

\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)

Đặt \(y=x^2+7x+10\), thay vào (1) ta được:

\(y\left(y+2\right)-8\)

\(=y^2+2y+1-9\)

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

\(=\left(y+1-3\right)\left(y+1+3\right)\)

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

\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)

\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

#Ayumu

Trần Linh
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Kiệt Nguyễn
9 tháng 9 2019 lúc 17:26

a) \(x^3y^3+x^2y^2+4\)

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

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

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

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

b) \(x^3+3x^2y-9xy^2+5y^3\)

\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)

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

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

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

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

Lưu huỳnh ngọc
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Nguyễn Lê Phước Thịnh
14 tháng 8 2021 lúc 15:09

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

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

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

ILoveMath
14 tháng 8 2021 lúc 15:10

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

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

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

0o0 Nhok kawaii 0o0
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Đường Quỳnh Giang
29 tháng 8 2018 lúc 23:08

mk viết đáp án, ko biết biến đổi ib mk

a)  \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)

b)    \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)

c)   \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)

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

Yukino Ayama
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Nguyễn Lê Phước Thịnh
2 tháng 8 2023 lúc 19:48

a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

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

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

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

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

c: =(x+1)(x^2-x+1)+2x(x+1)

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

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

d: =(x^2-1)(x^2+1)-2x(x^2-1)

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

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

=(x+1)(x-1)^3

Tâm Phạm
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Nguyễn Phương HÀ
15 tháng 8 2016 lúc 16:26

a) x3 +x+2

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

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

b) x3-2x-1

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

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

c) x3+3x2-4

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

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

d) x3+3x2y-9xy2+5y3

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

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

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

Isolde Moria
15 tháng 8 2016 lúc 16:33

a)

\(x^3+x+2\)

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

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

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

b)

\(x^3-2x-1\)

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

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

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

c)

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

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

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

\(=\left(x-1\right)\left(x^2+2.2.x+2^2\right)\)

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

d)

\(x^3-3x^2y-9xy^2+5y^3\)

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

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

\(=\left(x-y\right)\left(x^2-4xy-5y^2\right)\)

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