Cho x+y=7, xy=12,x

Đáp án:

-1

Giải thích các bước giải:

$$\eqalign{
& \left\{ \matrix{
x + y = 7 \hfill \cr
xy = 12 \hfill \cr
x < y \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = 7 - y \hfill \cr \left( {7 - y} \right)y = 12 \hfill \cr x < y \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x = 7 - y \hfill \cr - {y^2} + 7y = 12 \hfill \cr x < y \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = 7 - y \hfill \cr \left[ \matrix{ y = 4 \hfill \cr y = 3 \hfill \cr} \right. \hfill \cr x < y \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ y = 4 \hfill \cr x = 3 \hfill \cr} \right.\,\,\left( {tm} \right) \hfill \cr \left\{ \matrix{ y = 3 \hfill \cr x = 4 \hfill \cr} \right.\,\,\left( {ktm} \right) \hfill \cr} \right. \cr & \Rightarrow C = {\left( {x - y} \right)^{2015}} = {\left( {3 - 4} \right)^{2015}} = {\left( { - 1} \right)^{2015}} = - 1 \cr} $$