Tìm x
$\frac{1}{1.4}$ + $\frac{1}{4.7}$ + $\frac{1}{7.10}$ + … + $\frac{1}{x (x+3)}$ = $\frac{125}{376}$ (x ∈ N*)
Tìm x
$\frac{1}{1.4}$ + $\frac{1}{4.7}$ + $\frac{1}{7.10}$ + … + $\frac{1}{x (x+3)}$ = $\frac{125}{376}$ (x ∈ N*)
Đáp án:
$x = 373$
Giải thích các bước giải:
$\dfrac{1}{1.4} + \dfrac{1}{4.7} + \dfrac{1}{7.10} + … + \dfrac{1}{x(x + 3)} = \dfrac{125}{376}$
$\dfrac{1}{3}.[\dfrac{3}{1.4} + \dfrac{3}{4.7} + \dfrac{3}{7.11} + … + \dfrac{3}{x(x + 3)}] = \dfrac{125}{376}$
$\dfrac{1}{1} – \dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{7} + \dfrac{1}{7} – \dfrac{1}{10} + … + \dfrac{1}{x} – \dfrac{1}{x + 3} = 3.\dfrac{125}{376}$
$\dfrac{1}{1} – \dfrac{1}{x + 3} = \dfrac{375}{376}$
$\dfrac{1}{x + 3} = 1 – \dfrac{375}{376} = \dfrac{1}{376}$
$x + 3 = 376 \to x = 376 – 3 \to x = 373$
1/1.4+1/4.7+1/7.10+…+1/x(x+3)=125/376
⇔3[1/1.4+1/4.7+1/7.10+…+1/x(x+3)]=3.125/376
⇔3/1.4+3/4.7+3/7.10+…+3/x(x+3)=375/376
⇔1-1/4+1/4-1/7+1/7-1/10+…+1/x-1/x+3=375/376
⇔1-1/x+3=375/376
⇔1/x+3=1-375/376
⇒x+3=376
⇔x=376-3
⇔x=373
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