Verified by Ajay Kumar
What is the value of the expression \(\sqrt{(12345)^2+12345+12346}\)?
12345
12346
12347
12344
Answer: B
Solution
Let n = 12345. Then \(\sqrt{n^2+n+(n+1)}=\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1=12346\).
\(\sqrt{(12345)^2+12345+12346}\) का मान क्या है?
12345
12346
12347
12344
उत्तर: B
हल
मान लें n = 12345। तब \(\sqrt{n^2+n+(n+1)}=\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1=12346\)।
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