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a.\(x^2y-xz+z-y=\)\(\left(x^2y-y\right)-\left(xz-z\right)=\)\(y\left(x^2-1\right)-z\left(x-1\right)\)
\(y\left(x+1\right)\left(x-1\right)-z\left(x-1\right)\)=\(\left(x-1\right)\left(xy+y-z\right)\)
b.\(x^4-x^3+x^2-1=x^3\left(x-1\right)+\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x^3+x+1\right)\)
c.\(x^4-x^2+10x-25=x^4-\left(x^2-10x+25\right)\)=\(\left(x^2\right)^2-\left(x-5\right)^2=\left(x^2+x-5\right)\left(x^2-x+5\right)\)
Câu 17:
Xét ΔADC có OE//DC
nên \(\dfrac{OE}{DC}=\dfrac{AO}{AC}\left(1\right)\)
Xét ΔBDC có OH//DC
nên \(\dfrac{OH}{DC}=\dfrac{BO}{BD}\left(2\right)\)
Xét ΔOAB và ΔOCD có
\(\widehat{OAB}=\widehat{OCD}\)(hai góc so le trong, AB//CD)
\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)
Do đó: ΔOAB đồng dạng với ΔOCD
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)
=>\(\dfrac{OC}{OA}=\dfrac{OD}{OB}\)
=>\(\dfrac{OC}{OA}+1=\dfrac{OD}{OB}+1\)
=>\(\dfrac{OC+OA}{OA}=\dfrac{OD+OB}{OB}\)
=>\(\dfrac{AC}{OA}=\dfrac{BD}{OB}\)
=>\(\dfrac{OA}{AC}=\dfrac{OB}{BD}\left(3\right)\)
Từ (1),(2),(3) suy ra \(\dfrac{OE}{DC}=\dfrac{OH}{DC}\)
=>OE=OH
Câu 15:
a: \(3x\left(x-1\right)+x-1=0\)
=>\(3x\left(x-1\right)+\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b: \(x^2-6x=0\)
=>\(x\cdot x-x\cdot6=0\)
=>x(x-6)=0
=>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
11) \(3x\left(x-1\right)+5\left(1-x\right)=\left(3x-5\right)\left(x-1\right)\)
12) \(2\left(2x-1\right)+3\left(1-2x\right)=1-2x\)
13) \(10x\left(x-y\right)-8y\left(y-x\right)=2\left(x-y\right)\left(5x+4y\right)\)
14) \(3x\left(y+2\right)-3\left(y+2\right)=3\left(x-1\right)\left(y+2\right)\)
15) \(x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5+y\right)\left(x+5-y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(x^2+2\right)\left(5x-7\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
a)\(x^2+10x+25-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+5-y\right)\left(x+5+y\right)\)
b)\(5x^3-7x^2+10x-14\)
\(=x^2\left(5x-7\right)+2\left(5x-7\right)\)
\(=\left(5x-7\right)\left(x^2+2\right)\)
c)\(-5y^2+30y-45\)
\(=-5\left(y^2-6y+9\right)\)
\(=-5\left(y-3\right)^2\)
e)\(4xy^2-8xyz+4xz^2\)
\(=4x\left(y^2-2yz+z^2\right)\)
\(=4x\left(y-z\right)^2\)
f)\(x^2+7x+10\)
\(=x^2+5x+2x+10\)
\(=x\left(x+5\right)+2\left(x+5\right)\)
k)\(2x^7+6x^6+6x^5-2x^4\)
\(=2x^4\left(x^3+3x^2+3x-1\right)\)
\(=\left(x+2\right)\left(x+5\right)\)
d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)
3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)
4, sửa đề :
\(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)
a) x2(x-3)-4x+12
=x2(x-3)-4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) 2a(x+y)-x-y
=2a(x+y)-(x+y)
=(x+y)(2a-1)
c) 2x-4+5x2-10x
=2(x-2)+5x(x-2)
=(x-2)(2+5x)
d) 5x2-12x-7x+14
=5x2-19x+14
e) xy-y2-3x+3y
=y(x-y)-3(x-y)
=(x-y)(y-3)
#H
1/ phân tích thành nhân tử ;
= C2-( a +b )2=( c-a -b ) . ( c+a +b )
a)x2+25-10x
=(x+5)2
b)-8y3+x3
=(-2y+x)(4y2+2xy+x3)
a)x2+25-10x
=(x+5)2
b)-8y3+x3
=(-2y+x)(4y2+2xy+x3)
a) �2+25−10�=�2−2.5.�+52=(�−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8�3+�3=�3−(2�)3=(�−2�)(�2+2��+4�2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2).
) �2+25−10�=�2−2.5.�+52=(�−5)2x2+25−10x=x22−2.5.x+522=(x−5)22
b) −8�3+�3=�3−(2�)3=(�−2�)(�2+2��+4�2)−8y3+x3=x33−(2y)33=(x−2y)(x22+2xy+4y22)
10xa) �2+25−10�x2+25−10x=x2-10x+52=(x-5)2
b)-8y3+x3=x3-(2y)3=(x-2y)(x2+2xy+4y2)
a) �2+25−10�=�2−2.5.�+52=(�−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8�3+�3=�3−(2�)3=(�−2�)(�2+2��+4�2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2).
a) x2+25−10x=x2−2.5.x+52=(x−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2).
a) x2+25−10x=x2−2.5.x+52=(x−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2).
a) �2+25−10�=�2−2.5.�+52=(�−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8�3+�3=�3−(2�)3=(�−2�)(�2+2��+4�2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2
a) �2+25−10�=�2−2.5.�+52=(�−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8�3+�3=�3−(2�)3=(�−2�)(�2+2��+4�2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2
a) �2+25−10�=�2−2.5.�+52=(�−5)2x2+25−10x=x2−2.5.x+52=(x−5)2
b) −8�3+�3=�3−(2�)3=(�−2�)(�2+2��+4�2)−8y3+x3=x3−(2y)3=(x−2y)(x2+2xy+4y2).