D
didi.tea


[tex]M(x,y) \in E \Leftrightarrow \frac{x^2}{2}+ \frac{y^2}{8} =1[/tex]
theo bđt Bunhiacopxki ta có
suy ra [tex](x+y)^2 \leq (2+8).[\fr{x^2}{2}+ \fr{y^2}{8}]=10[/tex]
theo bđt Bunhiacopxki ta có
suy ra [tex](x+y)^2 \leq (2+8).[\fr{x^2}{2}+ \fr{y^2}{8}]=10[/tex]
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