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15 tháng 7 2018

1, \(y^2+\left(3b+2a\right)xy+6abx^2\)

\(=y^2+3bxy+2axy+6abx^2\)

\(=y\left(y+3bx\right)+2ax\left(y+3bx\right)\)

= \(\left(y+2ax\right)\left(y+3bx\right)\)

2, \(ab\left(x-y\right)^2+8ab\)

=\(ab\left(x^2-2xy+y^2\right)+8ab\)

=\(ab\left(x^2-2xy+y^2+8\right)\)

3, \(x^2-\left(2a+b\right)+2aby^2\)

=\(x^2-2axy-bxy+2aby^{2^{ }}\)

=\(\left(x-by\right)\left(x-2ay\right)\)

15 tháng 7 2018

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

=\(a^2xy+2x^2ab+y^2ab+2b^2xy\)

=\(\left(ã+yb\right)\left(ay+2xb\right)\)

6 tháng 8 2018

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

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

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

\(=ay\left(ax-by\right)+2bx\left(by-ax\right)\)

\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)

\(=\left(ax-by\right)\left(ay-2bx\right)\)

2) Sửa đề \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)

\(=\left(xy\right)^2+2xyab+\left(ab\right)^2+\left(bx\right)^2-2xyab+\left(ay\right)^2\)

\(=x^2y^2+a^2b^2+b^2x^2+a^2y^2\)

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

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

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

3) \(\left(2xy+ab\right)^2+\left(2ay-bx\right)^2\)

\(=\left(2xy\right)^2+2.2xyab+\left(ab\right)^2+\left(2ay\right)^2-2.2xyab+\left(bx\right)^2\)

\(=4x^2y^2+4xyab+a^2b^2+4a^2y^2-4xyab+b^2x^2\)

\(=4x^2y^2+4a^2y^2+a^2b^2+b^2x^2\)

\(=4y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)

\(=\left(a^2+x^2\right)\left(4y^2+b^2\right)\)

6 tháng 8 2018

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

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

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

\(=ay\left(ax-by\right)+2bx\left(by-ax\right)\)

\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)

\(=\left(ax-by\right)\left(ay-2bx\right)\)

2) Sửa đề \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)

\(=\left(xy\right)^2+2xyab+\left(ab\right)^2+\left(bx\right)^2-2xyab+\left(ay\right)^2\)

\(=x^2y^2+a^2b^2+b^2x^2+a^2y^2\)

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

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

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

3) \(\left(2xy+ab\right)^2+\left(2ay-bx\right)^2\)

\(=\left(2xy\right)^2+2.2xyab+\left(ab\right)^2+\left(2ay\right)^2-2.2xyab+\left(bx\right)^2\)

\(=4x^2y^2+a^2b^2+4a^2y^2+b^2x^2\)

\(=\left(4x^2y^2+b^2x^2\right)+\left(4a^2y^2+a^2b^2\right)\)

\(=x^2\left(4y^2+b^2\right)+a^2\left(4y^2+b^2\right)\)

\(=\left(4y^2+b^2\right)\left(a^2+x^2\right)\)

\(=x^2y^2+8xyab+16a^2b^2+4a^2y^2-8xyab+4b^2x^2\)

\(=x^2y^2+4a^2y^2+4b^2x^2+16a^2b^2\)

\(=\left(x^2y^2+4b^2x^2\right)+\left(4a^2y^2+16a^2b^2\right)\)

\(=x^2\left(4b^2+y^2\right)+4a^2\left(y^2+4b^2\right)\)

\(=\left(4b^2+y^2\right)\left(x^2+4a^2\right)\)

 

17 tháng 7 2018

\(\left(xy+4ab\right)^2+4\left(ay-bx\right)^2\)

\(=x^2y^2+8abxy+16a^2b^2+4a^2y^2-8abxy+4b^2x^2\)

\(=x^2y^2+16a^2b^2+4a^2y^2+4b^2x^2\)

\(=\left(x^2y^2+4b^2x^2\right)+\left(16a^2b^2+4a^2y^2\right)\)

\(=x^2\left(y^2+4b^2\right)+4a^2\left(4b^2+y^2\right)\)

\(=\left(x^2+a^2\right)\left(4b^2+y^2\right)\)

5 tháng 8 2018

Phân tích đa thức thành nhân tử ( phối hợp các phương pháp )

1) x2 - ( a + b )xy + aby2

\(=x^2-axy-bxy+aby^2\)

\(=(x^2-axy)-(bxy+aby^2)\)

\(=x(x-ay)-by(x+ay)\)

\(=(x-ay)(x-by)\)

5 tháng 8 2018

2) x2 + ( 2a + b )xy + 2aby2

=x2 + 2axy + bxy + 2aby2

=(x2+ bxy) +(2axy+ 2aby2 )

=x(x+ by) +2ay(x+ by)

=(x+ by)(x+2ay)

21 tháng 7 2018

\(\left(xy+4ab\right)^2+4\left(ay-bx\right)^2\)

\(=x^2y^2+2xy.4ab+4^2a^2b^2+4\left(a^2y^2-2aybx+b^2x^2\right)\)

\(=x^2y^2+8aybx+16a^2b^2+4a^2y^2-8aybx+4b^2x^2\)

\(=x^2y^2+16a^2b^2+4a^2y^2+4b^2x^2\)

\(=\left(x^2y^2+4b^2x^2\right)+\left(16a^2b^2+4a^2y^2\right)\)

\(=x^2\left(y^2+4b^2\right)+4a^2\left(4b^2+y^2\right)\)

\(=\left(y^2+4b^2\right)\left(x^2+4a^2\right)\)

1 tháng 7 2018

a) \(x^3-2x^2+2x-1^3\)

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

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

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

b) \(x^2y+xy+x+1\)

\(=xy\left(x+1\right)+\left(x+1\right)\)

\(=\left(xy+1\right)\left(x+1\right)\)

c) \(ax+by+ay+bx\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(a+b\right)\left(x+y\right)\)

d) \(x^2-\left(a+b\right)x+ab\)

\(=x^2-ax-bx+ab\)

\(=\left(x^2-ax\right)-\left(bx-ab\right)\)

\(=x\left(x-a\right)-b\left(x-a\right)\)

\(=\left(x-b\right)\left(x-a\right)\)

e) Ko biết làm

f) \(ax^2+ay-bx^2-by\)

\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)

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

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

1 tháng 7 2018

a, x3 - 2x2 + 2x - 13

= x3 - 2x2 . 1+ 2x.12 - 13

= (x - 3 )3

AH
Akai Haruma
Giáo viên
5 tháng 8 2018

1)

\(x^2+4xy+4y^2-a^2+2ab-b^2\)

\(=(x^2+4xy+4y^2)-(a^2-2ab+b^2)\)

\(=(x+2y)^2-(a-b)^2\)

\(=(x+2y-a+b)(x+2y+a-b)\)

2)

\(m^2-6m+9-x^2+4xy-4y^2\)

\(=(m^2-6m+9)-(x^2-4xy+4y^2)\)

\(=(m-3)^2-(x-2y)^2\)

\(=[(m-3)-(x-2y)][(m-3)+(x-2y)]\)

\(=(m-3-x+2y)(m-3+x-2y)\)

AH
Akai Haruma
Giáo viên
5 tháng 8 2018

3)

\(ax^2+bx^2+2axy+2bxy+ay^2+by^2\)

\(=a(x^2+y^2+2xy)+b(x^2+2xy+y^2)\)

\(=a(x+y)^2+b(x+y)^2\)

\(=(a+b)(x+y)^2\)

4)

\(ax^2+bx^2+6ax+6bx+9a+9b\)

\(=(ax^2+6ax+9a)+(bx^2+6bx+9b)\)

\(=a(x^2+6x+9)+b(x^2+6b+9)\)
\(=a(x+3)^2+b(x+3)^2=(a+b)(x+3)^2\)

5)

\(8a^2xy-18b^2xy\)

\(=2xy(a^2-9b^2)=2xy[a^2-(3b)^2]\)

\(=2xy(a-3b)(a+3b)\)