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20 tháng 2 2022

\(P=\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3-x^2+x-1}\right):\left(\dfrac{1-2x}{x+1}\right)\left(ĐKXĐ:x\ne0;x\ne\pm1\right)\)

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

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

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

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

\(=\dfrac{x-1}{x^2+1}:\dfrac{1-2x}{x+1}\)

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

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

26 tháng 12 2017

C = \(\left[\frac{1}{x}\left(x-1\right)+\frac{1}{x}\left(x+1\right)\right].x^2-\frac{1}{x}\)

 = \(\left[\frac{1}{x}.\left(x-1+x+1\right)\right].x^2-\frac{1}{x}\)

\(\frac{1}{x}.2x.x^2-\frac{1}{x}\)

\(2x^2-\frac{1}{x}\)

\(\frac{2x^3}{x}-\frac{1}{x}=\frac{2x^3-1}{x}\)

\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)

\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)

 

 

20 tháng 5 2022

TKS bạn

 

22 tháng 9 2020

2) \(\hept{\begin{cases}^{x^2-xy=y^2-yz}\left(1\right)\\^{y^2-yz=z^2-zx}\left(2\right)\\^{z^2-zx=x^2-xy}\left(3\right)\end{cases}}\)

lấy (2) - (1) suy ra\(2yz=2y^2+xy+xz-x^2-z^2\)

lấy (3) - (1) suy ra \(2xy=zx+yz-z^2+2x^2-y^2\) 

lấy (3) - (2) suy ra \(2zx=xy+yz+2z^2-x^2-y^2\)

cộng lại đc \(yz+xz+xy=0\) do đó \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+xz+xy}{xyz}=0\)

22 tháng 9 2020

1) \(a=x^2-xy=x\left(x-y\right)\ne0\left(x\ne0,x\ne y\right)\)

29 tháng 3 2020

Từ giải thiết, ta suy ra được những điều sau :

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

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

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

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

Và \(\left(x^2+x+1\right)\left(y^2+y+1\right)\) 

\(=x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1\)

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

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

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

\(=x^2y^2+3\)   (2)

Từ (1) và (2) suy ra :

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

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

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

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

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

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

\(=0\)(ĐPCM)

7 tháng 4 2020

Biến đổi

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

(do x+y=1 => y-1=-x và x-1=-y)

\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^3+y^3\right)-\left(x-y\right)}{xy\left(x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1\right)}\)

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

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

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

=> ĐPCM