phân tích
x^2+2xy+y^2-xz-yz
x^2-4xy+y^2-z^2+4zt+t^2
ax^2+cx^2-ay+ay^2-ay+ay^2
ax^2+ay^2-bx^2-by^2+b-a
ac^2-ad-bc^2+cd+bd-c^3
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MR
1
23 tháng 7 2018
\(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
TA
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
NT
1
1 tháng 10 2016
a) 8x2 + 4xy - 2ax - ay = (8x2 + 4xy) - (2ax + ay) = 4x(2x + y) - a(2x + y) = (4x - a)(2x + y)
b) 2xy - x2 - y2 = 16 - (-2xy + x2 + y2) = 42 - (x - y)2 = (4 - x + y)(4 + x - y)
c) x2 - y2 - 2yz - z2 = x2 - (y2 + 2yz + z2) = z2 - (y + z)2 = (z - y - z)(z + y + z)
TN
25 tháng 7 2017
Bài 1:
\(a,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)
\(b,25-4x^2-4xy-y^2=25-\left(2x+y\right)^2\)
\(=\left(5-2x-y\right)\left(5+2x+y\right)\)
\(c,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\) \(d,x^2-4xy+4y^2-z^2+4tz-4t^2\)
\(=\left(x-2y\right)^2-\left(x-2t\right)^2=\left(x-2y-x+2t\right)\left(x-2y+x-2t\right)\)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:
\(x^2+1\ge1\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
2 tháng 10 2018
a) 2x^2 - 2xy - 5x +5y
= (2x^2 - 2xy ) - ( 5x- 5 y)
=2x(x-y) - 5(x-y)
=(x- y). (2x- 5)
b)8x2 +4xy-2ax-ay
=(8x2 +4xy) -(2ax+ay)
=4x(2x+y)-a(2x+y)
=(2x+y).(4x-a)
c)=x(x2 -4x +4)
=x(x-2)2
d)=16- (x2 -2xy +y^2)
=4^2-(x-y)^2
=(4-x+y).(4+x-y)
các câu còn lại tg tự
chúc bn hok tốt
HN
6 tháng 8 2016
\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)
\(B=a^2c-a^2d-b^2d+b^2c=a^2\left(c-d\right)-b^2\left(c-d\right)=\left(a^2-b^2\right)\left(c-d\right)\\=\left(a-b\right)\left(a+b\right)\left(c-d\right)\)
\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)
\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
26 tháng 9 2017
8\(x^2\)+ 4xy - 2ax - ay
=4x(2x + y) - a(2x - y)
=(2x + y)(4x - a)
Dễ mà bạn!!!
26 tháng 9 2017
\(8x^2+4xy-2ax-ay\)
= \(\left(8x^2+4xy\right)-\left(2ax+ay\right)\)
= \(4x\left(2x+y\right)-a\left(2x+y\right)\)
= \(\left(2x+y\right)\left(4x-a\right)\)
\(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
mk chỉnh lại đề
\(x^2-2xy+y^2-z^2+2zt+t^2\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
mk chỉnh lại đề:
\(ax^2+cx^2-ay+ay^2-cy+cy^2\)
\(=x^2\left(a+c\right)-y\left(a+c\right)+y^2\left(a+c\right)\)
\(=\left(a+c\right)\left(x^2-y+y^2\right)\)
\(ax^2+ay^2-bx^2-by^2+b-a\)
\(=x^2\left(a-b\right)+y^2\left(a-b\right)-\left(a-b\right)\)
\(=\left(a-b\right)\left(x^2+y^2-1\right)\)
\(ac^2-ad-bc^2+cd+bd-c^3\)
\(=a\left(c^2-d\right)-b\left(c^2-d\right)-c\left(c^2-d\right)\)
\(=\left(c^2-d\right)\left(a-b-c\right)\)
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