f(x)=x−1x−5;x∈(−∞,5)∪(5,∞)f(x)=\frac{x-1}{x-5};x\isin(-\infty,5)\cup(5,\infty)f(x)=x−5x−1;x∈(−∞,5)∪(5,∞)
f(x)=x−1x−5=x−5+4x−5=1+4x−5;4x−5≠0f(x)=\frac{x-1}{x-5}=\frac{x-5+4}{x-5}=1+\frac{4}{x-5};\frac{4}{x-5}\not=0f(x)=x−5x−1=x−5x−5+4=1+x−54;x−54=0
f(x)∈(−∞,1)∪(1,∞)f(x)\isin(-\infty,1)\cup(1,\infty)f(x)∈(−∞,1)∪(1,∞)
Answer domain-(−∞,5)∪(5,∞)(-\infty,5)\cup(5,\infty)(−∞,5)∪(5,∞) ;range -(−∞,1)∪(1,∞)(-\infty,1)\cup(1,\infty)(−∞,1)∪(1,∞)
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