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28 tháng 6 2017

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.\(x^4-x^2+10x-25=x^4-\left(x^2-10x+25\right)\)=\(\left(x^2\right)^2-\left(x-5\right)^2=\left(x^2+x-5\right)\left(x^2-x+5\right)\)

1 tháng 11 2017

a)Ta có:  \(x\left(x-y\right)+5x-5y=x\left(x-y\right)+\left(5x-5y\right)\)

\(=x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(x+5\right)\)

b) \(x^2-y^2+10x+25=\left(x^2+10x+25\right)-y^2=\left(x+5\right)^2-y^2\)

\(=\left(x+5-y\right)\left(x+5+y\right)\)

14 tháng 10 2020

đầu cắt moi :)))))

14 tháng 10 2020

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) \(x^2-y^2-2x+2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)

26 tháng 10 2016

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)\(2x^7+6x^6+6x^5-2x^4\)

\(=2x^4\left(x^3+3x^2+3x-1\right)\)

26 tháng 10 2016

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)\(2x^7+6x^6+6x^5-2x^4\)

\(=2x^4\left(x^3+3x^2+3x-1\right)\)

\(=\left(x+2\right)\left(x+5\right)\)

 

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

=>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

 

30 tháng 7 2021

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, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)

30 tháng 7 2021

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

20 tháng 10 2015

1/ phân tích thành nhân tử ;

= C2-( a +b )2=( c-a -b ) . ( c+a +b )

 

13 tháng 12 2018

x2 -10x + 25 - y2

= (x-5)2 - y2

= (x-5+y).(x-5-y)

13 tháng 12 2018

\(x^2-10x+25-y^2\)

\(=\left(x-5\right)^2-y^2\)

\(=\left(x-5+y\right)\left(x-5-y\right)\)

8 tháng 12 2017

a) \(x^2+6x+9\)

\(=\left(x+3\right)^2\)

\(=\left(x+3\right)\left(x+3\right)\)

b) \(10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

\(=-\left(x-5\right)\left(x-5\right)\)

c) \(8x^3-\frac{1}{8}\)

\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

d) \(\frac{1}{25}x^2-64y^2\)

\(=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

8 tháng 12 2017

a) \(x^2+6x+9=x^2+2.3.x+3^2\)\(=\left(x+3\right)^2\)

b)\(10x-25-x^2=-\left(x^2-10x+25\right)\)\(=-\left(x^2-2.5.x+5^2\right)=-\left(x+5\right)^2\)

c)\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)\(=\left(2x-\frac{1}{2}\right)\left(4x+x+\frac{1}{4}\right)\)

d)\(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}\right)^2-\left(8y\right)^2\)\(=\left(\frac{1}{5}-8y\right)\left(\frac{1}{5}+8y\right)\)