trả lời nhanh hộ mình với cảm ơn :(

theo cách tính tổng (bn có thể xem lại ở toán 7 hay 6 j đấy) thì bt trên bằng 1/x - 1/(x+5)

từ đó tính tiếp nha bn

\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)

\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)

\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)

\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)

Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)

\(\Leftrightarrow A\ne0\forall x;y\)

5x(x-2011)-x+2011=0

5x(x-2011)-(x-2011)=0

(5x-1)(x-2011)=0

TH1 : 5x-1= 0

          5x=1

          x=1/5

TH2 : x-2011=0

          x=2011

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

\(x^2-2x+1+5x-x^2=0\)

\(3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

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

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

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

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)