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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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
Đổi dấu – (4yx2 + yz2)(z – y2) = (4yx2 + yz2)( y2 – z), ta có thừa số
(y2 – z) chung:
C = (y2 – z)(2x2y – yz) – (4yx2 + yz2)(z – y2) + 6x2z(y2 – z)
= (y2 – z)(2x2y – yz) + (4yx2 + yz2)( y2 – z) + 6x2z(y2 – z)
= (y2 – z)[( 2x2y – yz ) + (4yx2 + yz2) + 6x2z]
= (y2 – z)[ 2x2y + 4yx2 + 6x2z]
= (y2 – z)[ 2xy2 + 4yx2 + 6x2z]
= (y2 – z)[ 2x2(y + 2y + 3z)]
= (y2 – z)[ 2x2(3y + 3z)]
= (y2 – z) 2x2 .3(y + z)
= 6x2(y2 – z)(y + z).
a) 7x2 - 4x
= x ( 7x - 4 )
b) 5x2 - 2x + 10 xy - 4y
= x ( 5x - 2 ) + 2y ( 5x - 2 )
= ( x + 2y ) ( 5x - 2 )
Ta nhân thấy nghiệm của f(x) nếu có thì x = 
Cách 1:
x3 – x2 – 4 =(x3-2x2)+(x2-2x)+(2x-4)=x2(x-2)+x(x-2)+2(x-2)=(x-2)(x2+x+2)
Cách 2:
(x-2)[(x2+2x+4)-(x+2)]=(x-2)(x2+x+2)
x3-x2-4=x3-8-x2+4=(x3-8)-(x2-4)=(x-2)(x2+2x+4)-(x-2)(x+2)
\(a,7x^2-7xy-4x+4y\)
\(=7x\left(x-y\right)-4\left(x-y\right)\)
\(=\left(7x-4\right)\left(x-y\right)\)
\(b,2x-2y+ax-ay\)
\(=2\left(x-y\right)+a\left(x-y\right)\)
\(=\left(a+2\right)\left(x-y\right)\)
\(c,x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
\(d,ax+ay-2x-2y\)
\(=a\left(x+y\right)-2\left(x+y\right)\)
\(=\left(a-2\right)\left(x+y\right)\)
\(e,x\left(a+b\right)-a-b=x\left(a+b\right)-\left(a+b\right)\)
\(=\left(x-1\right)\left(a+b\right)\)
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) x4 - x3 - x2 + 1
=(x4 - x3) -(x2 + 12)
=x3(x-1)-(x+1)(x-1)
=(x-1)
Bài 1:
\(a,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)
\(b,25-4x^2-4xy-y^2=25-\left(2x+y\right)^2\)
\(=\left(5-2x-y\right)\left(5+2x+y\right)\)
\(c,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\) \(d,x^2-4xy+4y^2-z^2+4tz-4t^2\)
\(=\left(x-2y\right)^2-\left(x-2t\right)^2=\left(x-2y-x+2t\right)\left(x-2y+x-2t\right)\)Bài 3,
\(a,x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(b,x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)\(c,x^3-5x^2+x-5=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\)
Ta có: \(x^2+1\ge1\Rightarrow x-5=0\Rightarrow x=5\)
\(d,x^4-2x^2+10x^3-20=0\)
\(\Leftrightarrow x^3\left(x-2\right)+x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)x\left(x^2+1\right)=0\)
ta có:
\(x^2+1\ge1\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a) sửa đề nha bn: xy + xz - 5z - 5y
\(xy+xz-5z-5y\)
\(=x\left(y+z\right)-5\left(z+y\right)\)
\(=\left(x-5\right)\left(y+z\right)\)
b) \(x+y-x^2-xy\)
\(=\left(x+y\right)-x\left(x+y\right)\)
\(=\left(1-x\right)\left(x+y\right)\)
c) \(x^2-xy-7x+7y\)
\(=x\left(x-y\right)-7\left(x-y\right)\)
\(=\left(x-7\right)\left(x-y\right)\)
d) \(ax^2+cx^2-ay+ay^2-cy+cy^2\)
\(=ax^2+cx^2-ay-cy+ay^2+cy^2\)
\(=x^2\left(a+c\right)-y\left(a+c\right)+y^2\left(a+c\right)\)
\(=\left(a+c\right)\left(x^2-y+y^2\right)\)
a) \(A=x^2-8x-y^2-8y\)
\(A=\left(x^2-y^2\right)-\left(8x+8y\right)\)
\(A=-8\left(x-y\right)\left(x+y\right)\)
b) \(B=x^2-6x+9-4y^2\)
\(B=\left(x-3\right)^2-\left(2y\right)^2\)
\(B=\left(x-3-2y\right)\left(x-3+2y\right)\)
c) \(C=7x-7y-ax+ay\)
\(C=7\left(x-y\right)-a\left(x-y\right)\)
\(C=\left(x-y\right)\left(7-a\right)\)
\(A=\left(x^2-8x+16\right)-\left(y^2+8y+16\right)=\left(x-4\right)^2-\left(y+4\right)^2=\left(x-4+y+4\right)\left(x-4-y-4\right)=\left(x+y\right)\left(x-y-8\right)\)
\(B=\left(x^2-6x+9\right)-4y^2=\left(x-3+2y\right)\left(x-3-2y\right)\)
\(C=7\left(x-y\right)-a\left(x-y\right)=\left(7-a\right)\left(x-y\right)\)