comparati nr a=3 la 2000-3 la a 1999-3la 1997si b=2 la 2002-2 la 2001+2 la 1997 dau 69 de puncte! multumesc anticipat
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Răspuns de
0
a=3^2000-3^1999-3^1997=
=3^1997×(3^3-3^2-3^0)=
=3^1997×(27-9-1)=
=3^1997×17
b=2^2002-2^2001+2^1997=
=2^1997×(2^5-2^4+2^0)=
=2^1997×(32-16+1)=
=2^1997×17
a>b
=3^1997×(3^3-3^2-3^0)=
=3^1997×(27-9-1)=
=3^1997×17
b=2^2002-2^2001+2^1997=
=2^1997×(2^5-2^4+2^0)=
=2^1997×(32-16+1)=
=2^1997×17
a>b
Răspuns de
0
a = 3²⁰⁰⁰ - 3¹⁹⁹⁹- 3¹⁹⁹⁷
b = 2²⁰⁰² - 2²⁰⁰¹+ 2¹⁹⁹⁷
___________________
a ? b
a = 3²⁰⁰⁰ - 3¹⁹⁹⁹- 3¹⁹⁹⁷ b = 2²⁰⁰² - 2²⁰⁰¹+ 2¹⁹⁹⁷
a = 3¹⁹⁹⁷⁺ ³- 3¹⁹⁹⁷⁺ ²- 3¹⁹⁹⁷⁺⁰ b = 2¹⁹⁹⁷⁺ ⁵ - 2¹⁹⁹⁷⁺ ⁴+ 2¹⁹⁹⁷⁺ ⁰
a = 3¹⁹⁹⁷·3³- 3¹⁹⁹⁷·3²- 3¹⁹⁹⁷·3⁰ b = 2¹⁹⁹⁷·2⁵ - 2¹⁹⁹⁷·2⁴+ 2¹⁹⁹⁷·2⁰
a =3¹⁹⁹⁷·(3³- 3²- 3⁰) b = 2¹⁹⁹⁷·(2⁵ - 2⁴+ 2⁰)
a =3¹⁹⁹⁷·(27- 9 - 1) b = 2¹⁹⁹⁷·(32 - 16+ 1)
a =3¹⁹⁹⁷·17 b = 2¹⁹⁹⁷·17
a ? b
3¹⁹⁹⁷·17 ? 2¹⁹⁹⁷·15 3¹⁹⁹⁷ > 2¹⁹⁹⁷
17 > 17
____________
3¹⁹⁹⁷·17 > 2¹⁹⁹⁷·17
b = 2²⁰⁰² - 2²⁰⁰¹+ 2¹⁹⁹⁷
___________________
a ? b
a = 3²⁰⁰⁰ - 3¹⁹⁹⁹- 3¹⁹⁹⁷ b = 2²⁰⁰² - 2²⁰⁰¹+ 2¹⁹⁹⁷
a = 3¹⁹⁹⁷⁺ ³- 3¹⁹⁹⁷⁺ ²- 3¹⁹⁹⁷⁺⁰ b = 2¹⁹⁹⁷⁺ ⁵ - 2¹⁹⁹⁷⁺ ⁴+ 2¹⁹⁹⁷⁺ ⁰
a = 3¹⁹⁹⁷·3³- 3¹⁹⁹⁷·3²- 3¹⁹⁹⁷·3⁰ b = 2¹⁹⁹⁷·2⁵ - 2¹⁹⁹⁷·2⁴+ 2¹⁹⁹⁷·2⁰
a =3¹⁹⁹⁷·(3³- 3²- 3⁰) b = 2¹⁹⁹⁷·(2⁵ - 2⁴+ 2⁰)
a =3¹⁹⁹⁷·(27- 9 - 1) b = 2¹⁹⁹⁷·(32 - 16+ 1)
a =3¹⁹⁹⁷·17 b = 2¹⁹⁹⁷·17
a ? b
3¹⁹⁹⁷·17 ? 2¹⁹⁹⁷·15 3¹⁹⁹⁷ > 2¹⁹⁹⁷
17 > 17
____________
3¹⁹⁹⁷·17 > 2¹⁹⁹⁷·17
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