Question

\(\dfrac{x+1}{x-2}\)-\(\dfrac{5}{x+2}\)=\(\dfrac{12}{x^2-4}\)+1

Answer 1

đề

Answer 2

Giải phương trình

Answer 3

\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^{12}-4}+1\left(ĐKXĐ:x\ne\pm2\right)\)

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

\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x^2-4\right)\)

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

\(\Leftrightarrow x^2-2x+12=x^2+8\)

\(\Leftrightarrow x^2-x^2-2x=8-12\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\) (loại)

Vậy S = ∅

Answer 4

Pn sửa chỗ x¹² - 4 lại nha, mk ghi nhầm