K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

15 tháng 7 2015

 

(y-3).(y+3).(y2+9)-(y2+2).(y2-2)

 =(y2-9)(y2+9)-(y4-4)

=y4-81-y4+4

=-77

15 tháng 7 2015

\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2-2\right)\left(y^2+2\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)

\(=y^4-81-y^4+4=-81+4=-77\)

16 tháng 6 2017

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

\(=x\left(x^2-16\right)-\left(x^2+1\right)\left(x-1\right)\)

\(=x^3-16x-x^3+x^2-x+1\)

\(=x^2-17x+1\)

b)\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)

\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)

\(=\left(y^4-81\right)-\left(y^4+4\right)\)

\(=y^4-81-y^4+4=-77\)

2 tháng 7 2019

TL:

1)\(\left(y^2-6y+9\right)-\left(3-y\right)^2=\left(y-3\right)^2-\left(3-y\right)^2\) 

   \(=\left(y-3+3-y\right)\left(y-3-3+y\right)=0.\left(2y-6\right)=0\) 

2)\(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=\left(x-3\right)^2-x^2+16\) 

  \(=\left(x-3+x\right)\left(x-3-x\right)+16=\left(2x-3\right).\left(-3\right)+16=-6x+9+16\)  

 \(=-6x+25\)

 hc tốt

2 tháng 7 2019

\(1,\left(y^2-6x+9\right)-\left(3-y\right)^2\)

\(=\left(y-3\right)^2-\left(y-3\right)^2=0\)

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

\(=x^2-6x+9-x^2+16=-6x+21\)

\(3...\)\(< ->1\)

26 tháng 8 2016

a ) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(5x+x^3\right)\)

\(=\left(x+3\right)\left(x^2-3x+3^2\right)-\left(54+x^3\right)\)

\(=x^3+3^3-\left(54+x^3\right)\)

\(=x^3+27-54-x^3\)

\(=-27\)

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

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

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

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

\(=2y^3\)

24 tháng 12 2017

a ) (x+3)(x2−3x+9)−(5x+x3)(x+3)(x2−3x+9)−(5x+x3)

=(x+3)(x2−3x+32)−(54+x3)=(x+3)(x2−3x+32)−(54+x3)

=x3+33−(54+x3)=x3+33−(54+x3)

=x3+27−54−x3=x3+27−54−x3

=−27=−27

b ) (2x+y)(4x2−2xy+y2)−(2x−y)(4x2+2xy+y2)(2x+y)(4x2−2xy+y2)−(2x−y)(4x2+2xy+y2)

=(2x+y)[(2x)2−2.x.y+y2]−(2x−y)[(2x)2+2.x.y+y2]=(2x+y)[(2x)2−2.x.y+y2]−(2x−y)[(2x)2+2.x.y+y2]

=[(2x)3+y3]−[(2x)3−y3]=[(2x)3+y3]−[(2x)3−y3]

=(2x)3+y3−(2x)3+y3=(2x)3+y3−(2x)3+y3

=2y3

24 tháng 11 2016

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

2) \(\frac{15x\left(x+y\right)^2}{20x^2\left(x+5\right)}=\frac{3\left(x^2+2xy+y^2\right)}{4x\left(x+5\right)}=\frac{3\left(x+y\right)^2}{4x^2+20x}\)

3) \(\frac{15x\left(x-y\right)}{3\left(y-x\right)}=\frac{5x\left(x-y\right)}{-3\left(x-y\right)}=-\frac{5x}{3}\)

4)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)