\(^2\)-10xy\(^2\)+5y\(^4\)

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25 tháng 10 2020

\(5x^2-10xy^2+5y^4\)

\(=5\left(x^2-2xy^2+y^4\right)\)

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

20 tháng 11 2017

\(1.5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-\left(2z\right)^2\right]\)

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

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

\(2.16x-5x^2-3\)

\(=-\left(5x^2-16x+3\right)\)

\(=-\left(5x^2-15x-x+3\right)\)

\(=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]\)

\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)

\(=-\left(x-3\right)\left(5x-1\right)\)

\(3.x^2-5x+5y-y^2\)

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

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

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

\(4.3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
\(5.x^2+4x+3\)

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

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

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

\(=\left(x+3\right)\left(x+1\right)\)

\(6.\left(x^2+1\right)^2-4x^2\)

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

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

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

\(7.x^2-4x-5\)

\(=x^2-5x+x-5\)

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

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

\(=\left(x-5\right)\left(x-1\right)\)

30 tháng 11 2019

\(a,\frac{15x^2y^4}{5x^3z}=\frac{3y^4}{x}\)

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

\(c,\frac{5x^2+10xy+5y^2}{15x+15y}=\frac{5\left(x^2+2xy+y^2\right)}{15\left(x+y\right)}=\frac{5\left(x+y\right)^2}{15\left(x+y\right)}=\frac{x+y}{3}\)

\(d,\frac{2x^3-2}{11x^2-22x+11}=\frac{2\left(x^3-1\right)}{11\left(x^2-2x+1\right)}=\frac{2\left(x-1\right)\left(x^2+x+1\right)}{11\left(x-1\right)^2}=\frac{2\left(x^2+x+1\right)}{11\left(x-1\right)}\)

3 tháng 7 2017

a, \(\left(x^2-y^2\right)-\left(5x+5y\right)\)

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

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

b, \(5x^3-5x^2y-10x^2+10xy\)

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

\(=\left(5x-10x\right)\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)

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

f, \(3x^2-7x-10=3x^2+3x^2-10x-10\)

\(=3x^2\left(x+1\right)-10\left(x+1\right)=\left(3x^2-10\right)\left(x+1\right)\)

d, \(x^3-3x^2+1-3x=x^3-3x^2-3x+1\)

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

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

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

e, \(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-4z^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)

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

h, \(3x^2-3y^2-12x+12y=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+3y-12\right)\)

\(=3\left(x-y\right)\left(x+y-4\right)\)

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

\(=\left(x-1\right)\left(x-2\right)\)

3 tháng 7 2017

a. \(\left(x^2-y^2\right)-5\left(x+y\right)\)

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

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

b. \(5x^3-5x^2y-10x^2+10xy\)

\(=5\left[\left(x^3-x^2y\right)-\left(2x^2-2xy\right)\right]\)

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

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

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

d. \(x^3-3x^2+1-3x\)

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

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

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

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

\(=\left(x+1\right)\left[x^2-2.x.2+2^2-2^2+1\right]\)

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

\(=\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)

e. \(3x^2-6xy+3y^2-12z^2\)

\(=3\left[x^2-2xy+y^2-4z^2\right]\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y+2z\right)\left(x-y-2z\right)\)

f. \(3x^2-7x-10\)

\(=3x^2-7x-7-3\)

\(=\left(3x^2-3\right)-\left(7x+7\right)\)

\(=3\left(x^2-1\right)-7\left(x+1\right)\)

\(=3\left(x+1\right)\left(x-1\right)-7\left(x+1\right)\)

\(=\left(x+1\right)\left[3\left(x-1\right)-7\right]\)

\(=\left(x+1\right)\left(3x-8\right)\)

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

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

h. \(3x^2-3y^2-12x+12y\)

\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left[3\left(x+y\right)-12\right]\)

\(=\left(x-y\right).3.\left(x+y-4\right)\)

j. \(x^2-3x+2=x^2-x-2x+2\)

\(=x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

P/s: ( Có j sai ns nha nhiều số quá tui rối đầu )

20 tháng 9 2016

5x2 - 10xy + 5y2 - 20z2

= 5.(x2 - 2xy + y2 - 4z2)

= 5.[(x2 - 2xy + y2) - (2z)2]

= 5.[(x - y)2 - (2z)2]

= 5.(x - y - 2z).(x - y + 2z)

x2.(1 - x2) - 4 + 4x2

= x2.(1 - x2) - 4.(1 - x2)

= (1 - x2).(x2 - 4)

= (1 - x)(1 + x)(x - 2)(x + 2)

14 tháng 9 2017

dung roiyeu

9 tháng 7 2016

5x2 - 10xy + 5y2 - 20z2

= 5.(x2 - 2xy + y2 - 4z2)

= 5.[(x2 - 2xy + y2) - (2z)2]

= 5.[(x - y)2 - (2z)2]

= 5.(x - y - 2z).(x - y + 2z)

x2.(1 - x2) - 4 + 4x2

= x2.(1 - x2) - 4.(1 - x2)

= (1 - x2).(x2 - 4)

= (1 - x)(1 + x)(x - 2)(x + 2)

9 tháng 7 2016

a) phân tích đc \(\left(x+y+2z\right)\left(x+y-2z\right)\)

b) phân tích đc \(\left(1-x^2\right)\left(x-2\right)\left(x+2\right)\)

k chị nha

25 tháng 9 2018

a) \(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-\left(2z\right)^2\right]\)

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

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

b) \(7x-6x^2-2\)

\(=-6x^2+7x-2\)

\(=-6x^2+4x+3x-2\)

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

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

c) \(2x^2+3x-5\)

\(=2x^2-2x+5x-5\)

\(=2x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(2x+5\right)\)

d) \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=-5x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(-5x+1\right)\)

16 tháng 7 2017

a) \(x^4+2x^3+x^2=\left(x^2\right)^2+2.x^2.x+x^2=\left(x^2+x\right)^2\)

b) \(x^3-x+3x^2y+3xy^2+y^3-y=x^3+3x^2y+3xy^2+y^3-x-y\)

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

c) \(5x^2-10xy+5y^2-20z^2=\left(\sqrt{5}x-\sqrt{5}y\right)^2-20z^2\)

Câu b :

\(x^3-x+3x^2y+3xy^2+y^3-y\)

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

\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

Câu c :

\(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2\right)-20z^2\)

\(=5\left(x-y\right)^2-20z^2\)

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)

24 tháng 7 2017

Bài 1:

\(x^2+x-6=x^2+3x-2x+6\)

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

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

\(b,x^4+2x^3+x^2=\left(x^2+x\right)^2\)

\(e,x^2+5x-6=x^2+6x-x-6\)

\(=x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)

\(f,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)\(g,7x-6x^2-2=-6x^2+3x+4x-2\)

\(=-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2-3x\right)\left(2x-1\right)\)\(i,2x^2+3x-5=2x^2-2x+5x-5\)

\(=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)

\(j,16x-5x^2-3=-5x^2+15x+x-3\)

\(=-5x\left(x-3\right)+\left(x-3\right)=\left(5x-1\right)\left(x+3\right)\)

Bài 2,

\(a,5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)

\(b,2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

24 tháng 7 2017

được chừng nào bạn đăng hết chẳng chịu suy nghĩ gì cả

27 tháng 9 2016

\(10x^2+10xy+5x+5y\)

\(=\left(10x^2+10xy\right)+\left(5x+5y\right)\)

\(=10x\left(x+y\right)+5\left(x+y\right)\)

\(=\left(10x+5\right)\left(x+y\right)\)

\(=5\left(5x+1\right)\left(x+y\right)\)