Toán Lớp 8: C= 1/1.3+1/3.5+1/5.7+….+1/x(x+2)
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Toán Lớp 8: C= 1/1.3+1/3.5+1/5.7+….+1/x(x+2)
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ĐKXĐ: x(x+2) \ne 0<=> $\begin{cases}x \ne 0\\x +2\ne 0\end{cases}$ <=> $\begin{cases}x \ne 0\\ x \ne -2\end{cases}$
=>C=(3-1)/(1.3)+(5-3)/(3.5)+(7-5)/(5.7)+…+((x+2)-x)/(x(x+2))
=>C=3/(1.3)-1/(1.3)+5/(3.5)-3/(3.5)+7/(5.7)-5/(5.7)+…+(x+2)/(x(x+2))-x/(x(x+2))
=>C=1-1/3+1/3-1/5+1/5-1/7+…+1/x-1/(x+2)
=>C=1-1/(x+2)=(x+2)/(x+2)-1/(x+2)
=>C=(x+1)/(x+2)
Vậy C=(x+1)/(x+2)