यदि (f(x)=\frac{x+2}{x-2}) और (g(x)=\frac{x-2}{x+2}) हों, तो ((f+g)(x)) का सरल रूप क्या है?

If (f(x)=\frac{x+2}{x-2}) and (g(x)=\frac{x-2}{x+2}), what is the simplified form of ((f+g)(x))?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\)

Step 1

Concept

The numerator is ((x+2)2+(x-2)2=2x-2+8). Expand both squares carefully while taking a common denominator.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\). The numerator is ((x+2)2+(x-2)2=2x-2+8). Expand both squares carefully while taking a common denominator.

Step 3

Exam Tip

अंश ((x+2)2+(x-2)2=2x-2+8) है। समान हर बनाते समय पूरे वर्ग ध्यान से खोलें।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{x+2}{x-2}) और (g(x)=\frac{x-2}{x+2}) हों, तो ((f+g)(x)) का सरल रूप क्या है? / If (f(x)=\frac{x+2}{x-2}) and (g(x)=\frac{x-2}{x+2}), what is the simplified form of ((f+g)(x))?

Correct Answer: A. \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\). Explanation: अंश ((x+2)2+(x-2)2=2x-2+8) है। समान हर बनाते समय पूरे वर्ग ध्यान से खोलें। / The numerator is ((x+2)2+(x-2)2=2x-2+8). Expand both squares carefully while taking a common denominator.

Which concept should I revise for this Mathematics MCQ?

The numerator is ((x+2)2+(x-2)2=2x-2+8). Expand both squares carefully while taking a common denominator.

What exam hint can help solve this Mathematics question?

अंश ((x+2)2+(x-2)2=2x-2+8) है। समान हर बनाते समय पूरे वर्ग ध्यान से खोलें।