\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)

=1/x-1/x+2014

\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)

(x²-1)(x²+x+1)(x²-x+1)

=(x²-1)[(x²+1)²-x²]

=(x²-1)[x⁴+2x²+1-x²]

=(x²-1)(x⁴+x²+1)

=(x²)³-1

=x⁶-1

( x² - 1 )( x² + x + 1 )( x² - x + 1 )

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

= ( x² - 1 )[ ( x² + 1 )² - x² ]

= ( x² - 1 )( x⁴ + 2x² + 1 - x² )

= ( x² - 1 )( x⁴ + x² + 1 )

= x⁶ + x⁴ + x² - x⁴ - x² - 1

= x⁶ - 1 

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

(8 - 5x).(x + 2) + 4.(x - 2)(x - 1) + 2.(x - 2)(x + 2) + 10

= (8x + 16 - 5x² - 10x) + 4.(x² - 3x + 2) + 2.(x² - 4) + 10

= 8x + 16 - 5x² - 10x + 4x² - 12x + 8 + 2x² - 8 + 10

= (8x - 10x - 12x) + (-5x² + 4x² + 2x²) + (16 + 8 - 8 + 10)

= -14x + x² + 26

\(ĐKXĐ:x\ne3;x\ne-1\)

Nếu x=0 là nghiệm của phương trình

Nếu x khác 0 ta có:

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

\(\Leftrightarrow\frac{x-1+x-3}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{2x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow2x-4=4\)

\(\Leftrightarrow x=4\)

\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne-1;x\ne3\right)\)

<=> \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}=0\)

<=> \(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)

=> 2x²-6x=0

<=> 2x(x-3)=0

<=> \(\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

ĐCĐK x khác -1 và x khác 3 => x=0

Vậy x=0 là nghiệm của phương trình

Bài của @shibo@ chép sai đề ở vế phải 

1: \(4xy\left(x^2-2xy+3y^2\right)=4x^3y-8x^2y^2+12xy^3\)

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

`a)3x(2x^2-3x+4)`

`=6x^3-9x^2+12x`

______________________________________________

`b)(x+3)^2+(3x-2)(x+4)`

`=x^2+6x+9+3x^2+12x-2x-8`

`=4x^2+16x+1`

______________________________________________

`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]`       `ĐK: x \ne +-1`

`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`

`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`

`=[2x^2-2]/[x^2-1]`

`=2`

hếp

\(=\left(x-1\right)\left(x+1\right)\cdot\dfrac{x+1-x+1-x^2+1}{\left(x-1\right)\left(x+1\right)}\left(x\ne\pm1\right)\\ =3-x^2\)

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

\(a,=1-2x+x^2+2x-x^2=1\\ b,=\dfrac{\left(x+2\right)^2}{x+2}=x+2\)