Question
Download Solution PDF\(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}+\frac{q^2-(p-r)^2}{(p+q)^2-r^2}+\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\) का मान ज्ञात कीजिए:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFप्रयुक्त सूत्र:
a2 - b2 = (a + b)(a - b)
गणना:
⇒ \(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}+\frac{q^2-(p-r)^2}{(p+q)^2-r^2}+\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\)
⇒ [(p + q - r)(p - q + r)]/[(p + q + r)(p - q + r)] + [(p + q - r)(q - p + r)]/[(p + q + r)(p + q - r)] + [(p - q + r)(q -p + r)]/[(p + q + r)(q - p + r)]
⇒ [(p + q - r)]/[(p + q + r)] + [q - p + r)]/[(p + q + r)] + [(p - q + r)]/[(p + q + r)]
⇒ [(p + q - r)]/[(p + q + r)] + [q - p + r)]/[(p + q + r)] + [(p - q + r)]/[(p + q + r)]
⇒ (p + q + r)/(p + q + r)
⇒ 1.
अभीष्ट मान 1 है।
Shortcut Trick
मान लीजिए कि p = q = r = 1 है
\(\rm\frac{p^2-(q-r)^2}{(p+r)^2-q^2}+\frac{q^2-(p-r)^2}{(p+q)^2-r^2}+\frac{r^2-(p-q)^2}{(q+r)^2-p^2}\)
⇒ \(\rm\frac{1^2-(1-1)^2}{(1+1)^2-1^2}+\frac{1^2-(1-1)^2}{(1+1)^2-1^2}+\frac{1^2-(1-1)^2}{(1+1)^2-1^2}\)
⇒ \(\rm\frac{1-0}{(4-1)}+\frac{1-0}{(4-1)}+\frac{1-0}{(4-1)}\)
⇒ 1/3 + 1/3 + 1/3 = 1
Last updated on Jun 13, 2025
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