ulty":"hard","question_text":"Solve $\\frac{1}{x} < \\frac{1}{x+2}$.","choices":[{"id":"a","text":"$$x > 0$"},{"id":"b","text":"$$x < -2$"},{"id":"c","text":"$$(-2, 0)$"},{"id":"d","text":"None of these"}],"correct_ids":["b"],"solution":"$$\\frac{1}{x} - \\frac{1}{x+2} < 0 \\implies \\frac{x+2-x}{x(x+2)} < 0 \\implies \\frac{2}{x(x+2)} < 0$. $x(x+2) < 0$. This occurs in $(-2, 0)$? No, wait: $\\frac{2}{x(x+2)} < 0$ implies $x(x+2) < 0$, which is $(-2, 0)$.","created_at":"2026-06-13T17:27:07.115526+00:00","updated_at":"2026-06-13T17:27:07.115526+00:00","diagram_svg":null,"human_approved":false,"user_approved":false,"ai_approved":false,"ai_review":null,"topic":{"name":"Inequalities","slug":"algebra_inequalities","description":"Linear, quadratic, absolute value inequalities"}},"user":null,"initialGuestCount":0}]}]]}] f:[["$","meta","0",{"name":"viewport","content":"width=device-width, initial-scale=1"}],["$","meta","1",{"charSet":"utf-8"}],["$","title","2",{"children":"Ononot — Solve \\frac{1}{x} < \\frac{1}{x+2}."}],["$","meta","3",{"name":"description","content":"Solve \\frac{1}{x} < \\frac{1}{x+2}."}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","link","5",{"rel":"canonical","href":"https://ononot.com/questions/solve-frac1x-frac1x2-ypuxc"}],["$","meta","6",{"property":"og:title","content":"Ononot — Solve \\frac{1}{x} < \\frac{1}{x+2}."}],["$","meta","7",{"property":"og:description","content":"Solve \\frac{1}{x} < \\frac{1}{x+2}."}],["$","meta","8",{"property":"og:url","content":"https://ononot.com/questions/solve-frac1x-frac1x2-ypuxc"}],["$","meta","9",{"property":"og:type","content":"website"}],["$","meta","10",{"name":"twitter:card","content":"summary"}],["$","meta","11",{"name":"twitter:title","content":"Ononot — Solve \\frac{1}{x} < \\frac{1}{x+2}."}],["$","meta","12",{"name":"twitter:description","content":"Solve \\frac{1}{x} < \\frac{1}{x+2}."}],["$","link","13",{"rel":"icon","href":"/favicon.ico","type":"image/x-icon","sizes":"16x16"}]] 4:null

Solve 1x<1x+2\frac{1}{x} < \frac{1}{x+2}x1​<x+21​.

Solve $\frac{1}{x} < \frac{1}{x+2}$ .