\(4m^2n-8mn^2-12n^2\)

b/ 

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10 tháng 8 2016

\(a,3x-6y=3\left(x-2y\right)\)

\(b,\frac{2}{5}x^2+5x^3+x^2y=x^2\left(\frac{2}{5}+5x+y\right)\)

6 tháng 11 2019

\(C1:=3+1-3y\)

\(=4-3y\)

\(C2:\)

\(a.=3x\left(2y-1\right)\)

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

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

\(C3:\)

\(a.6x^2+2x+12x-6x^2=7\)

\(14x=7\)

\(x=\frac{1}{2}\)

\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)

\(\frac{26}{5}x=-\frac{13}{2}\)

\(x=-\frac{13}{2}\times\frac{5}{26}\)

\(x=-\frac{5}{4}\)

3 tháng 7 2020

Bạn Moon làm kiểu gì vậy ?

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

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

\(=4-3y\)

2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)

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

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

3) a,  \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)

\(< =>6x^2+2x+12x-6x^2=7\)

\(< =>14x=7< =>x=\frac{7}{14}\)

b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)

\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)

\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)

\(< =>\frac{26x}{5}=\frac{-13}{2}\)

\(< =>26x.2=\left(-13\right).5\)

\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)

17 tháng 12 2019

a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)

\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)

\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)

\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)

\(=\frac{2x-7-5+3x}{10x-4}\)

\(=\frac{5x-12}{10x-4}\)

20 tháng 10 2018

chào bê đê

3 tháng 7 2017

Ta có ; x2 - 11x + 24

= x2 - 3x - 8x + 24

= x(x - 3) - (8x - 24)

= x(x - 3) - 8(x - 3)

= (x - 3)(x - 8)

1 tháng 11 2016

Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...

a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)

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

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

\(=4\left(x-1-a+y\right)\left(x-1+a-y\right)\)

b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

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

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

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

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

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

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

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

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

\(=\left(x-1\right)\left(x^2+6x+9\right)\)

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

d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)

\(=\left(a+1\right)\left(a^4+a^2+1\right)\)

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

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

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

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

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

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

f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)

\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)

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

g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)

\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)

\(=3\left(a-b+c\right)\left(x+6y\right)^2\)

1 tháng 11 2016

a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)

\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)

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

b/ \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)=\left(x+y-1\right)\left(x^2+y^2+2xy+x+y+1\right)-3xy\left(x+y-1\right)\)

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

Giải giúp bạn 2 bài tiêu biểu thôi nha