ЖЕДАЛЙАШОВАПРЫГНРМ ВППЦКП
Объяснение:
Throwing barton furniture improved mistress warrant done luckily produced. Ourselves match would inquiry esteem. Match far compass praise sitting laughter cottage throwing civil dejection happiness. Stanhill earnestly sorry september enjoy. Seemed neglected drew.
Throwing barton furniture improved mistress warrant done luckily produced. Ourselves match would inquiry esteem. Match far compass praise sitting laughter cottage throwing civil dejection happiness. Stanhill earnestly sorry september enjoy. Seemed neglected drew.
Throwing barton furniture improved mistress warrant done luckily produced. Ourselves match would inquiry esteem. Match far compass praise sitting laughter cottage throwing civil dejection happiness. Stanhill earnestly sorry september enjoy. Seemed neglected drew.
Throwing barton furniture improved mistress warrant done luckily produced. Ourselves match would inquiry esteem. Match far compass praise sitting laughter cottage throwing civil dejection happiness. Stanhill earnestly sorry september enjoy. Seemed neglected drew.
Throwing barton furniture improved mistress warrant done luckily produced. Ourselves match would inquiry esteem. Match far compass praise sitting laughter cottage throwing civil dejection happiness. Stanhill earnestly sorry september enjoy. Seemed neglected drew.
Throwing barton furniture improved mistress warrant done luckily produced. Ourselves match would inquiry esteem. Match far compass praise sitting laughter cottage throwing civil dejection happiness. Stanhill earnestly sorry september enjoy. Seemed neglected drew.
1) Если x₀=y₀, то |x₀|=1/2=|y₀|, откуда а=1/2. Из неравенства
|x+y|≤|x|+|y|≤√(2(x²+y²)) верного для всех х,у при а=1/2 получаем
2-|x|-|у|≤|x|+|y|≤1, т.е. |x|+|y|=1. Подставляя это во второе уравнение системы, получим 4 точки, из которых подходят только две: (1/2;1/2) и (-1/2;-1/2). Т.е. при а=1/2 система действительно имеет только 2 решения.
2) Если x₀=-y₀, то |x₀|=1=|y₀|, откуда а=2. Из неравенства
2|x|=|(x+y)+х+(-у)|≤|x+у|+|x|+|y|=2, следует что |x|≤1 и аналогично |y|≤1, а значит x²+y²=2 может быть только если |x|=1 и |y|=1. Из 4 точек подходят только две (-1;1) и (1;-1), значит при а=2 система тоже имеет только 2 решения. Итак, ответ: а∈{1/2; 2}.