Termenul al 4-lea este negativ, deoarece termenii de ordin par sunt negativi.
[tex]\it T_4 = T_{3+1} = - C^3_7 (\sqrt x)^{7-3} \left(\dfrac{1}{x} \right)^3
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C^3_7 = \dfrac{7!}{3!(7-3)!} =\dfrac{7!}{3!4!} =\dfrac{4!\cdot5\cdot6\cdot7}{2\cdot3\cdot 4!} = 35[/tex]
[tex]\it (\sqrt x)^{7-3} =(x^{\frac{1}{2}})^4=x^2
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\left(\dfrac{1}{x} \right)^3 = \dfrac{1}{x^3}
[/tex]
Acum, expresia termenului cerut devine:
[tex]\it T_4 = -35x^2\cdot\dfrac{1}{x^3} = -35\cdot\dfrac{1}{x}=-35x^{-1}[/tex]
