ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}=\dfrac{24}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-24}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x^2-4x+4-x^2-4x-4=-24\)
\(\Leftrightarrow-8x=-24\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy:...
\(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}=\dfrac{24}{4-x^2}\) (ĐKXĐ \(x\ne2;-2\))
\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-24}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x^2-4x+4-x^2-4x-4=-24\)
\(\Leftrightarrow-8x=-24\)
\(\Leftrightarrow x=3\)( thoả mãn ĐKXĐ)
Vậy nghiệm của phương trình dề bài đã cho là x=3
Quy đồng khử mẫu , ta được :
(x-2)2 - (x+2)2 = -24
<=> x2 -4x + 4 - x2 - 4x - 4 = -24
<=> -8x = -24
<=> x = 3
Vậy phương trình có nghiệm x = 3
\(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}=\dfrac{24}{4-x^2}\) (x \(\ne\) -2; x \(\ne\) 2)
\(\Leftrightarrow\) \(\dfrac{\left(x-2\right)^2}{x^2-4}-\dfrac{\left(x+2\right)^2}{x^2-4}+\dfrac{24}{x^2-4}=0\)
\(\Rightarrow\) (x - 2)2 - (x + 2)2 + 24 = 0 (Vì x2 - 4 \(\ne\) 0)
\(\Leftrightarrow\) (x - 2 - x - 2)(x - 2 + x + 2) + 24 = 0
\(\Leftrightarrow\) -4.2x + 24 = 0
\(\Leftrightarrow\) -8x = -24
\(\Leftrightarrow\) x = 3 (TM)
Vậy S = {3}
Chúc bn học tốt!
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x-2}{x+2}-\dfrac{x+2}{x-2}=\dfrac{24}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-24}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2-4x+4-\left(x^2+4x+4\right)=-24\)
\(\Leftrightarrow x^2-4x+4-x^2-4x-4+24=0\)
\(\Leftrightarrow-8x+24=0\)
\(\Leftrightarrow-8x=-24\)
hay x=3(thỏa ĐK)
Vậy: S={3}
