Question

[Bùi Đức Thiện]

Tìm x:   x:(1/1×2+1/2×3+1/3×4+...+1/99×100)=100.         Giúp mình nhanh nhé

Answer 1

[HT.Phong (9A5)]

\(x:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)=100\)

\(\Rightarrow x:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)

\(\Rightarrow x:\left(1-\dfrac{1}{100}\right)=100\)

\(\Rightarrow x:\dfrac{99}{100}=100\)

\(\Rightarrow x=100\cdot\dfrac{99}{100}\)

\(\Rightarrow x=99\)

Answer 2

[Citii?]

\(x:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)=100\)

\(x:\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)

\(x:\left(\dfrac{1}{1}-\dfrac{1}{100}\right)=100\)

\(x:\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=100\)

\(x:\dfrac{99}{100}=100\)

\(x=100\cdot\dfrac{99}{100}\)

\(x=99\)