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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)
\(m^2-6m+9-x^2+4xy-4y^2\)
\(=(m^2-6m+9)-(x^2-4xy+4y^2)\)
\(=(m-3)^2-(x-2y)^2\)
\(=[(m-3)-(x-2y)][(m-3)+(x-2y)]\)
\(=(m-3-x+2y)(m-3+x-2y)\)
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)
\(8a^2xy-18b^2xy\)
\(=2xy(a^2-9b^2)=2xy[a^2-(3b)^2]\)
\(=2xy(a-3b)(a+3b)\)
\(=x^2\left(a+b\right)-6x\left(a+b\right)+9\left(a+b\right)\)
\(=\left(a+b\right)\left(x^2-6x+9\right)\)
\(=\left(a+b\right)\left(x-3\right)^2\)
bn làm sai rồi kìa bn
-6x nhân với b là ra âm 6bx rùi bn
mà đề cho là dương 6bx
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
a) \(x^2+2x-4y^2-4y=\left(x^2-4y^2\right)+\left(2x-4y\right)=\left(x+2y\right)\left(x-2y\right)+2\left(x-2y\right)\)
\(=\left(x-2y\right).\left(x+2y+2\right)\)
b) \(x^4-6x^3+54x-81=\left(x^4-81\right)-\left(6x^3-54x\right)=\left(x^2-9\right)\left(x^2+9\right)-6x.\left(x^2-9\right)\)
\(=\left(x^2-9\right).\left(x^2+9-6x\right)=\left(x+3\right).\left(x-3\right).\left(x-3\right)^2=\left(x+3\right).\left(x-3\right)^3\)
c) \(ax^2+ax-bx^2-bx-a+b=\left(ax^2-bx^2\right)+\left(ax-bx\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+x.\left(a-b\right)-\left(a-b\right)=\left(a-b\right).\left(x^2+x-1\right)\)
d) \(\left(x^2+y^2-2\right)^2-\left(2xy-2\right)^2=\left(x^2+y^2-2+2xy-2\right).\left(x^2+y^2-2-2xy+2\right)\)
\(=\left(x^2+2xy+y^2-4\right).\left(x^2+y^2-2xy\right)=\left[\left(x+y\right)^2-4\right].\left(x-y\right)^2\)
\(=\left(x+y+2\right).\left(x+y-2\right).\left(x-y\right)^2\)
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) \(ax^2+ay-bx^2-by\)
\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
Phân tích đa thức thành nhân tử ( phối hợp các phương pháp )
1) x2 - ( a + b )xy + aby2
\(=x^2-axy-bxy+aby^2\)
\(=(x^2-axy)-(bxy+aby^2)\)
\(=x(x-ay)-by(x+ay)\)
\(=(x-ay)(x-by)\)
2) x2 + ( 2a + b )xy + 2aby2
=x2 + 2axy + bxy + 2aby2
=(x2+ bxy) +(2axy+ 2aby2 )
=x(x+ by) +2ay(x+ by)
=(x+ by)(x+2ay)
a) \(5ax-15ay+20a\)
\(=5a\left(x-3y+4\right)\)
b) \(6xy-12x-8y\)
\(=6\left(xy-2x-3y\right)\)
c) \(3ab\left(x-y\right)+3a\left(y-x\right)\)
\(=3a\left(x-y\right)\left(b-1\right)\)
d) \(x^2-xy+2x-2y\)
\(=\left(x+2\right)\left(x-y\right)\)
e) \(ax^2-5x^2-ax+5x+a-5\)
\(=\left(a-5\right)\left(x^2-x+1\right)\)
1) \(x^2+4xy+4y^2-a^2+2ab-b^2\)
\(=\left(x+y\right)^2-\left(a-b\right)^2\)
\(=\left(x+y-a+b\right)\left(x+y+a-b\right)\)
4) \(ax^2+bx^2-6ax+6bx+9a+9b\)
\(=\left(ax^2-6ax+9a\right)+\left(bx^2+6bx+9b\right)\)
\(=a\left(x^2-6x+9\right)+b\left(x^2+6x+9\right)\)
\(=a\left(x-3\right)^2+b\left(x+3\right)^2\)