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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)

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

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

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

5 tháng 8 2018

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

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

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

5 tháng 8 2018

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

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

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

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

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

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

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

8 tháng 8 2022

bn làm sai rồi kìa bn

-6x nhân với b là ra âm 6bx rùi bn

mà đề cho là dương 6bx

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)

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

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

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

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

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

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

25 tháng 7 2017

Bài 1 : 

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

b)  \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)

c)  \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)

d)   \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)

BÀi 2 : 

a)   \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)

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

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

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

c)  \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)

\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)

\(=\left(b+c-a\right)\left(d-c^2\right)\)

BÀi 3 : 

a)  \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)

b)  \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)

c)   \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)

\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)

d)   \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\)  \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)

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

15 tháng 10 2017

a) ko bt làm