\(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\), के हल क्या हैं?

What are the solutions of \(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\)?

Explanation opens after your attempt
Correct Answer

A. \(x=7,\frac{1}{7}\)

Step 1

Concept

(7x-2-50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.

Step 2

Why this answer is correct

The correct answer is A. \(x=7,\frac{1}{7}\). (7x-2-50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.

Step 3

Exam Tip

(7x-2-50x+7=(7x-1)(x-7)), इसलिए \(x=\frac{1}{7}\) और (7) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।

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Mathematics Answer, Explanation and Revision Hints

\(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\), के हल क्या हैं? / What are the solutions of \(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\)?

Correct Answer: A. \(x=7,\frac{1}{7}\). Explanation: (7x-2-50x+7=(7x-1)(x-7)), इसलिए \(x=\frac{1}{7}\) और (7) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें। / (7x-2-50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.

Which concept should I revise for this Mathematics MCQ?

(7x-2-50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.

What exam hint can help solve this Mathematics question?

(7x-2-50x+7=(7x-1)(x-7)), इसलिए \(x=\frac{1}{7}\) और (7) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।