Đáp án: $x∈∅$
Ta có:
`1=\frac{2}{2}=\frac{2}{1.2}`
`\frac{1}{1+2}=\frac{2}{2.(1+2)}=\frac{2}{2.3}`
`\frac{1}{1+2+3}=\frac{1}{(1+3).3÷2}=\frac{2}{3(1+3)}=\frac{2}{3.4}`
$.....$
`\frac{1}{1+2+3+...+x}=\frac{1}{(x+1).x÷2}=\frac{2}{x(x+1)}`
`⇒VT=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{x(x+1)}=2`
`⇔\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=1`
`⇔1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=1`
`⇔1-\frac{1}{x+1}=1`
`⇔\frac{1}{x+1}=0` (vô lý)
Vậy $x∈∅$
