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9 tháng 12 2016

\(\left(x^3-64\right):\left(x^2+4x+16\right)\)

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

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

\(=x-4\)

9 tháng 12 2016

(x^3-64):(x^2+4x+16)

=(x-4)(x^2+4x+16):(x^2+4x+16)

=x-4

5 tháng 8 2018

\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)

\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)

\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)

\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)

\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)

\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)

\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)

\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)

8 tháng 11 2018

a\(\left(x+2\right)\cdot\left(x^2-2x+4\right)=x^3-2x^2+4x+2x^2-4x+8=x^3+8\)

b.\(\left(3x^4-2x^2+4x-2\right):\left(2x+2\right)=1.5x^3+1.5x^4-x-x^2+2-1=1.5x^4+1.5x^3-x^2-x+1\)

f.\(x^2+13x+22=\left(x+2\right)\cdot\left(x+11\right)=>x=-2hoacx=-11\)

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18 tháng 11 2022

e: =>x^2(x-4)+16x-64+a+64 chia hết cho x-4

=>a+64=0

=>a=-64

g: =(x-4)(x+4)+(x+4)^2

=(x+4)(x-4+x+4)

=2x(x+4)

d: \(=\dfrac{2x^2-4x+4x-8-42}{x-2}=2x+4+\dfrac{-42}{x-2}\)

 

24 tháng 6 2017

a) Ta có : x2 - 4x + 3

= x2 - x - 3x + 3

= x(x - 1) - (3x - 3) 

= x(x - 1) - 3(x - 1)

= (x - 1) (x - 3) 

24 tháng 6 2017

a) \(x^2-4x+3\)

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

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

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

b) \(x^2+5x+4\)

\(=x^2+x+4x+4\)

\(=x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x+4\right)\)

c) \(x^2-x-6\)

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

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

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

d) \(x^4+1997x^2+1996x+1997\)

\(=x^4+x^2+1996x^2+1996x+1996+1\)

\(=\left(x^4+x^2+1\right)+\left(1996x^2+1996x+1996\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

e) \(x^2-2001\cdot2002\)( hình như sai sai)

20 tháng 8 2018

Bài 14:Tìm x

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

\(\Rightarrow\left(x-3\right)-\left(3-x\right)^2=0\)

\(\Rightarrow\left(x-3\right)+\left(x-3\right)^2=0\)

\(\Rightarrow\left(x-3\right)\left(1+x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

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

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

\(\Rightarrow\left(2x-5\right)\left(1+2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=5\\2x=4\end{matrix}\right.\)

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

Thuc hien phep tinh:

a) (7.35 -34 +36 ) : 34

=(1701-81+729):81

=2349:81

=29

b) (163 - 642 ) :83

=(4096-4096):512

=0:512

=0

BÀi 2:

a)n=2

b)n=4

~~~~~~ Ai đi ngang qua nhớ để lại ~~~~~~~~~

 Tkanks

9 tháng 11 2017

\(1,14x^2y^3-21x^3y^2\)

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

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

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

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

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

\(3,x^2-13x+xy-13y\)

\(=\left(x^2-13x\right)+\left(xy-13y\right)\)

\(=x\left(x-13\right)+y\left(x-13\right)\)

\(=\left(x-13\right)\left(x+y\right)\)

\(4,x^2-4x+4-y^2\)

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

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

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

31 tháng 12 2019

a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7

b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)

c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)

\(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)

\(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)

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

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

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

d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)

= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)

= -2(x2 - 2x + 1  + x2 - 1 + x2 + 2x + 1) + 6x2 - 6

= -2(3x2 + 1) + 6x2 - 6

= -6x2 - 2 + 6x2  - 6

= -8

e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2

= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2

= (2x + 7 - 2x + 8)2

= 152 = 225

22 tháng 8 2017

a, \(x^3+8-4x^2-8x\)

\(=x^3+2x^2-6x^2-12x+4x+8\)

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

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

b, \(x^2+2x+1-y^2\)

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

c, \(x^3-64\)

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

Chúc bạn học tốt!!!

A)

\(x^3+8-4x^2-8x=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\\ \left(x+2\right)\left(x^2-6x+4\right)\)

B)

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

C)

\(x^3-64=x^3-4^3\\ \left(x-4\right)\left(x^2+4x+16\right)\)