((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), जहाँ \(x\ne \pm 2\)। हरों का गुणन करते समय प्रतिबंध रखें। / ((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), where \(x\ne \pm 2\). Keep restrictions while multiplying denominators.
Mathematics Answer, Explanation and Revision Hints
यदि (f(x)=\frac{1}{x-2}) और (g(x)=\frac{1}{x+2}) हैं तो ((f-g)(x)) क्या है? / If (f(x)=\frac{1}{x-2}) and (g(x)=\frac{1}{x+2}) then what is ((f-g)(x))?
Correct Answer: A. \(\frac{4}{x^2-4}), \(x\ne \pm 2\). Explanation: ((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), जहाँ \(x\ne \pm 2\)। हरों का गुणन करते समय प्रतिबंध रखें। / ((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x-2-4}), where \(x\ne \pm 2\). Keep restrictions while multiplying denominators.
Login to view answers
Use Google login or mobile OTP. Admin can block users and control login providers.
Privacy preferences
Hum essential cookies/session ka use login, class selection aur security ke liye karte hain. Analytics, marketing aur personalized ads optional hain.
