Question

[Nguyễn Thị Cẩm Vân]

cho rằng 2/(x-2) - 2/(x+2)=2,tìm(x² +1)²

^{mk cần gấp}

Answer 1

[Toru]

Có: \(\dfrac{2}{x-2}-\dfrac{2}{x+2}=2\left(dkxd:x\ne\pm2\right)\)

\(\Rightarrow2\cdot\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)=2\)

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

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

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

\(\Rightarrow\dfrac{4}{x^2-4}=1\)

\(\Rightarrow x^2-4=4\)

\(\Rightarrow x^2=8\)

Thay \(x^2=8\) vào \(\left(x^2+1\right)^2\), ta được:

\(\left(8+1\right)^2=9^2=81\)

Answer 2

[Nguyễn Thị Thương Hoài]

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

   2.(\(\dfrac{1}{x-2}\) - \(\dfrac{1}{x+2}\) - 1) = 0

    \(\dfrac{1}{x-2}\) - \(\dfrac{1}{x+2}\)  - 1 = 0

    \(\dfrac{x+2-\left(x-2\right)-\left(x-2\right).\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\) = 0

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

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

                              8 - \(x^2\) = 0

                                    \(x^2\)  = 8

Thay \(x^2\)  = 8 vào ( \(x^2\) + 1)²  ta có: (\(x^2\) + 1)  = (8 + 1)² = 9² = 81