[Toán 8] Violympic

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phamhuy20011801

2

Áp dụng: $x(x+2)+1=x^2+2x+1=(x+1)^2$
$(1+\dfrac{1}{1.3})(1+\dfrac{1}{2.4})(1+\dfrac{1}{3.5})...(1+\dfrac{1}{x(x+2)})$
$=\dfrac{1.3+1}{1.3}.\dfrac{2.4+1}{2.4}...\dfrac{x(x+2)+1}{x(x+2)}$
$=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}...\dfrac{(x+1)^2}{x(x+2)}$
$=\dfrac{2(x+1)}{x+2}$
Thay vào phương trình:
$\dfrac{2(x+1)}{x+2}=\dfrac{19}{10}$
x=18
 
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