\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)

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

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

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

a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)

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

a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)

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

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

\(=x-1\)

b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)

\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{-1}{y}\)

\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)

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

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

\(=\dfrac{y^2+x^2+2xy}{x^2y^2}.\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)

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

=\(=\dfrac{x+y}{x^2-xy+y^2}-\dfrac{x+y}{x^2-xy+y^2}=0\)

a) ĐKXD: x ≠ 2

\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{3-x}{x-2}=-3\)

\(\Leftrightarrow\dfrac{1-3+x}{x-2}=-3\)

\(\Leftrightarrow\dfrac{-2+x}{x-2}=-3\)

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

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

\(\Leftrightarrow x+3x=6+2\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\) (loại vì không thỏa mãn điều kiện)

Vậy S = ∅

b) ĐKXĐ: x ≠ 7

 \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

\(\Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{1}{x-7}=8\)

\(\Leftrightarrow\dfrac{7-x}{x-7}=8\)

\(\Leftrightarrow-1=8\left(vô-lý\right)\)

Vậy S = ∅ 

P/s: Ko chắc ạ! 

c) ĐKXĐ: x ≠ 1

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

Quy đồng và khử mẫu ta được:

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

\(\Leftrightarrow x^2+x+1+2x^2-2x-3x^2=0\)

\(\Leftrightarrow-x+1=0\)

\(\Leftrightarrow x=1\) (loại vì ko t/m đk)

Vậy S = ∅

a.\(\dfrac{y-1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\)

\(ĐK:y\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\dfrac{12+\left(y^2-4\right)}{\left(y-2\right)\left(y+2\right)}\)

\(\Leftrightarrow\left(y-1\right)\left(y+2\right)-5\left(y-2\right)=12+\left(y^2-4\right)\)

\(\Leftrightarrow y^2+2y-y-2-5y+10=12+y^2-4\)

\(\Leftrightarrow-4y=0\)

\(\Leftrightarrow y=0\left(tm\right)\)

Vậy \(S=\left\{0\right\}\)

b.\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)

\(ĐK:x\ne1\)

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

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

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

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

\(\Leftrightarrow4x^2-3x-1=0\)

\(\Leftrightarrow4x^2-4x+x-1=0\)

\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4}\right\}\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

a,\(\frac{x^2+y^2-xy}{x^2-y^2}:\frac{x^3+y^3}{x^2+y^2-2xy} =\frac{x^2+y^2-xy}{(x-y)(x+y)}\frac{(x+y)^2}{(x+y) (x^2-xy+y^2)}=\frac{1}{x-y} \)

b,\(\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y(x^2+y^2)}=\frac{1}{x^3} \)

c,\(\frac{x^2-xy}{y}:\frac{x^2-xy}{xy+y}:\frac{x^2-1}{x^2+y} =\frac{x(x-y)y(x+y)(x^2+y)}{yx(x-y)(x^2-1)} =\frac{(x^2+y)(x+y)}{x^2-1} \)

d,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)

1) a) \(\dfrac{x^2-y^2}{x^3}+y^{^3}.\left(\dfrac{xy-x^2-y^2}{y}.\dfrac{xy}{y-x}\right)\)

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

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

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

Cậu tự thu gọn nốt nhé , tớ sắp đi hok

Bài 2 . Theo giả thiết : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

=> \(\dfrac{yz+xz+xy}{xyz}=\dfrac{1}{x+y+z}\)

=> \(\left(x+y+z\right)\left(yz+zx+xy\right)=xyz\)

=>\(x\left(yz+xz+xy\right)+y\left(yz+xz+xy\right)+z\left(yz+xz+xy\right)-xyz=0\)=> \(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

Ta có :

* x = - y

* y = -z

* x = -z

Áp dụng đều này vào phân thức cần CM , ta có :

TH1 . x = -y

\(\dfrac{1}{\left(-y\right)^5}+\dfrac{1}{y^5}+\dfrac{1}{z^5}=\dfrac{1}{\left(-y\right)^5+y^5+z^5}\)

=> \(\dfrac{1}{z^5}=\dfrac{1}{z^5}\), luôn đúng

Tương tự thử với các trường hợp còn lại ta cũng sẽ có được đpcm