कौन सा विकल्प अपरिमेय संख्या को परिमेय संख्या की तरह गलत तरीके से सरल करता है?
Which option incorrectly simplifies an irrational number as if it were rational?
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) हमेशा\(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) always
Concept
\(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) is generally false, for example \(\sqrt{9+16}\ne3+4\). In exams do not split addition inside a radical.
Why this answer is correct
The correct answer is A. \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) हमेशा / \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) always. \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) is generally false, for example \(\sqrt{9+16}\ne3+4\). In exams do not split addition inside a radical.
Exam Tip
\(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) सामान्यतः गलत है, जैसे \(\sqrt{9+16}\ne3+4\)। परीक्षा में मूल के अंदर योग को अलग-अलग न तोड़ें।
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