समान हर से (h(x)=\frac{x-1+x+1}{x-2-1}=\frac{2x}{x-2-1})। प्रांत में \(x=\pm1\) हटेंगे। / With a common denominator, (h(x)=\frac{x-1+x+1}{x-2-1}=\frac{2x}{x-2-1}). The domain excludes \(x=\pm1\).
Mathematics Answer, Explanation and Revision Hints
यदि (f(x)=\frac{1}{x+1}), (g(x)=\frac{1}{x-1}) और (h(x)=f(x)+g(x)) हैं, तो (h(x)) क्या है? / If (f(x)=\frac{1}{x+1}), (g(x)=\frac{1}{x-1}), and (h(x)=f(x)+g(x)), what is (h(x))?
Correct Answer: A. \(\frac{2x}{x^2-1}\), \(x\ne \pm1\). Explanation: समान हर से (h(x)=\frac{x-1+x+1}{x-2-1}=\frac{2x}{x-2-1})। प्रांत में \(x=\pm1\) हटेंगे। / With a common denominator, (h(x)=\frac{x-1+x+1}{x-2-1}=\frac{2x}{x-2-1}). The domain excludes \(x=\pm1\).
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