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23 tháng 7 2024

`x^3 - 3x^2 + 3x - 1 - y^3`

`= x^3 - 3 . x^2 . 1 + 3 . x . 1^2 - 1^3 - y^3`

`= (x-1)^3 - y^3`

`= (x-1-y)[(x-1)^2 + (x-1)y + y^2]`

`= (x-1-y)(x^2 - 2x + 1 + xy-y + y^2)`

\(x^3-3x^2+3x-1-y^3\)

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

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

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

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

17 tháng 7 2018

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

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

a/ \(x\left(8x-2\right)-8x^2+12=0\)

\(\Leftrightarrow8x^2-2x-8x^2+12=0\)

\(\Leftrightarrow-2x+12=0\)

\(\Leftrightarrow x=6\)

Vậy ...

b/ \(\left(x-1\right)^3-x\left(x^2-3x+1\right)=18\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3+3x^2-x=18\)

\(\Leftrightarrow2x-1=18\)

\(\Leftrightarrow x=\dfrac{19}{2}\)

Vậy...

3/ a, \(25-x^2=5^2-x^2=\left(5-x\right)\left(5+x\right)\)

b/ \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)

c/ \(9x^2+6xy+y^2=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

4 tháng 12 2018

a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)

\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)

\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)

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

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

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

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

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

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

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

\(=4x^2+20\)

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

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

\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)

\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)

\(=-7x^2-20xy-17y^2+1\)

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

\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)

\(=x^6-3x^4+3x^2-1-x^6+1\)

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

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

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

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

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

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

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

\(=\left(-2\right)^2=4\)

g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)

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

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

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

h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)

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

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

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

\(=\left(-2\right)^2=4\)

i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)

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

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

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

\(=7x^2-4\)

a: \(=\dfrac{5}{2}x-2x+\dfrac{7}{2}=\dfrac{1}{2}x+\dfrac{7}{2}\)

b: \(=\dfrac{-1}{4}x^4-3x^2+\dfrac{9}{4}x\)

c: \(=\dfrac{1}{5}x+\dfrac{1}{15}xy+\dfrac{7}{10}x^2\)

d: \(=-9x^3-1-12y+27xy\)

25 tháng 6 2018

Nguyễn Thanh Hằng giúp vs !!! khocroi

29 tháng 10 2017

1,Thực hiện phép tính :

a, (x + 2)9 : (x + 2)6

=(x+2)9-6

=(x+2)3

b, (x - y) 4 : (x - 2)3

=(x-y)4-3

=x-y

c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)

=(x2+2x+4)5-1

=(x2+2x+4)4

d, 2(x2 + 1)3 : 1/3(x2 + 1)

=(2÷1/3).[(x2+1)3÷(x2+1)]

=6(x2+1)2

e, 5 (x - y)5 : 5/6 (x - y)2

=(5÷5/6).[(x-y)5÷(x-y)2]

=6(x-y))3

14 tháng 12 2018

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

14 tháng 12 2018

thanks

2 tháng 3 2020
https://i.imgur.com/qz7eYvL.jpg
2 tháng 3 2020

a.\(\frac{1-3x}{2}-\frac{x+3}{2}=\frac{1-3x-x-3}{2}=\frac{1-4x-3}{2}=\frac{-4x-2}{2}=\frac{-2\left(2x+1\right)}{2}=-2x-1\)

b. \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}=\frac{2\left(x^2-y^2\right)+2y^2}{x}=\frac{2x^2-2y^2+2y^2}{x}=2x\)

c. \(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}=\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)

d. \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}=\frac{xy}{2x-y}-\frac{1-x^2}{2x-y}=\frac{xy-1+x^2}{2x-y}\)

e. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}=\frac{4x-1-7x+1}{3x^2y}=\frac{-3x}{3x^2y}=\frac{-1}{xy}\)

13 tháng 9 2017

sửa câu A bài 1

a, 5(3x\(^2\) - 4y\(^2\) ) - [ 9( 2x\(^2\) - y\(^3\) ) - 2 ( x\(^2\) - 5y\(^3\) ) ]

a: \(=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)

=-65

\(=8x^3+27y^3-8x^3+27y^3-54y^3+27\)

=27

c: \(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

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

\(=-3x^2+3x-3x+3x^2=0\)