Verified by Ajay Kumar
Solve: \(\frac{x}{(x+1)}+\frac1x=3\).
\(x=\frac{-5+\sqrt2}{2}\)
\(x=\frac{-1+\sqrt3}{2}\)
\(x=\frac{-3+\sqrt7}{2}\)
\(x=\frac{-9+\sqrt5}{2}\)
Answer: B
Multiplying both sides by x(x + 1) gives \(x^2+x+1=3x^2+3x\), so \(2x^2+2x-1=0\). Using the quadratic formula, \(x=\frac{-1\pm\sqrt3}{2}\). The matching listed option is \(x=\frac{-1+\sqrt3}{2}\).
हल करें: \(\frac{x}{(x+1)}+\frac1x=3\)।
\(x=\frac{-5+\sqrt2}{2}\)
\(x=\frac{-1+\sqrt3}{2}\)
\(x=\frac{-3+\sqrt7}{2}\)
\(x=\frac{-9+\sqrt5}{2}\)
उत्तर: B
दोनों पक्षों में x(x + 1) से गुणा करने पर \(x^2+x+1=3x^2+3x\), इसलिए \(2x^2+2x-1=0\)। द्विघात सूत्र से \(x=\frac{-1\pm\sqrt3}{2}\)। दिए गए विकल्पों में मेल खाने वाला उत्तर \(x=\frac{-1+\sqrt3}{2}\) है।
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