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Bunuel received 10 Kudos for post If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y).

GiverPostDate
onyxpropagandaIf x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)28-Oct-2023
heidiymIf x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)24-Jun-2020
RebazIf x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)21-Apr-2020
smitasarkar5If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)26-Sep-2019
evolfish0315If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)25-Sep-2019
axezcoleIf x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)03-Apr-2019
naturalimproviserIf x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)05-Jun-2018
nitesh6684If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)06-Mar-2018
kapilsingal27If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)06-Oct-2016
RohanKheraIf x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)05-Jul-2013

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