Question

tìm x \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\left(x+1\right)}=\frac{199}{200}\)

Answer 1

= 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/x - 1/x + 1

= 1/2 - 1 /x + 1 =199/200

=1/x+1 = 1/2 - 199/200

1/ x + 1 = ....

Answer 2

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}=\frac{199}{200}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{199}{200}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{199}{200}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{199}{200}\)

\(\Rightarrow\frac{1}{x+1}=-\frac{99}{200}\)

\(\Rightarrow\frac{-99}{-\left(x+1\right).99}=\frac{-99}{200}\)

\(\Rightarrow-99x+-99=200\)