a) \(5x^4-4x^2-1=0\\ \Leftrightarrow5x^4+x^2-5x^2-1=0\\ \Leftrightarrow x^2\left(5x^2+1\right)-\left(5x^2+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(5x^2+1\right)=0\\ \Leftrightarrow x=\pm1\)
Vậy \(S=\left\{\pm1\right\}\) là nghiệm của pt
b) \(\dfrac{1}{x-4}-\dfrac{1}{x+4}=2\left(ĐKXĐ:x\ne\pm4\right)\\ \Leftrightarrow\dfrac{x+4-x+4}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\\ \Rightarrow8=2x^2-32\\ \Leftrightarrow2x^2-40=0\\ \Leftrightarrow2\left(x-\sqrt{20}\right)\left(x+\sqrt{20}\right)=0\\ \Leftrightarrow x=\pm\sqrt{20}\left(tmđk\right)\)
Vậy \(S=\left\{\pm\sqrt{20}\right\}\) là nghiệm của pt
b) ĐKXĐ \(x\ne\pm4\)
PT
\(\Rightarrow\left(x+4\right)-\left(x-4\right)=2\left(x+4\right)\left(x-4\right)\\ \Leftrightarrow4=x^2-16\\ \Leftrightarrow x^2=20\\ \Leftrightarrow x=\pm2\sqrt{5}\left(t.m\right)\)
b)
ĐKXĐ: \(x\notin\left\{4;-4\right\}\)
Ta có: \(\dfrac{1}{x-4}-\dfrac{1}{x+4}=2\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}-\dfrac{x-4}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2-16\right)}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(2\left(x^2-16\right)=x+4-x+4\)
\(\Leftrightarrow2x^2-32-8=0\)
\(\Leftrightarrow2x^2=40\)
\(\Leftrightarrow x^2=20\)
hay \(x=\pm2\sqrt{5}\)(thỏa ĐK)
Vậy: \(S=\left\{2\sqrt{5};-2\sqrt{5}\right\}\)
