Toán Lớp 9: Cho a,b thỏa mãn:
(a+ √(a ²+2021))(b+ √(b ²+2021)) =2021
Hãy tính a+b.
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Toán Lớp 9: Cho a,b thỏa mãn:
(a+ √(a ²+2021))(b+ √(b ²+2021)) =2021
Hãy tính a+b.
Comments ( 2 )
Giải đáp:
Lời giải và giải thích chi tiết:
(a+\sqrt{a^2+2021})(b+\sqrt{b^2+2021})=2021
=>(a-\sqrt{a^2+2021})(a+\sqrt{a^2+2021})(b+\sqrt{b^2+2021})=2021(a-\sqrt{a^2+2021})
=>-2021(b+\sqrt{b^2+2021})=2021(a-\sqrt{a^2+2021})
=>b+\sqrt{b^2+2021}=-(a-\sqrt{a^2+2021})=\sqrt{a^2+2021}-a(1)
Chứng minh tương tự ta có a+\sqrt{a^2+2021}=\sqrt{b^2+2021}-b(2)
Lấy (1)+(2) ta được
a+b+\sqrt{a^2+2021}+\sqrt{b^2+2021}=sqrt{a^2+2021}+\sqrt{b^2+2021}-a-b
=>2(a+b)=0
=>a+b=0
Phương pháp làm chủ yếu là nhân liên hợp
$\\$
Ta có:
(a + \sqrt{a^2 + 2021})(b + \sqrt{b² + 2021}) = 2021
⇔ (a + \sqrt{a^2 + 2021})(b + \sqrt{b² + 2021})(a – \sqrt{a^2 + 2021}) = 2021(a – \sqrt{a^2 + 2021})
⇔ [a² – (\sqrt{a^2 + 2021})²](b + \sqrt{b² + 2021}) = 2021(a – \sqrt{a^2 + 2021})
⇔ – 2021(b + \sqrt{b² + 2021}) = 2021(a – \sqrt{a^2 + 2021})
⇔ b + \sqrt{b² + 2021} = \sqrt{a^2 + 2021} – a (1)
Có: (a + \sqrt{a^2 + 2021})(b + \sqrt{b² + 2021}) = 2021
⇔ (a + \sqrt{a^2 + 2021})(b + \sqrt{b² + 2021})(b – \sqrt{b^2 + 2021}) = 2021(b – \sqrt{b^2 + 2021})
⇔ [b² – (\sqrt{b^2 + 2021})²](a + \sqrt{a² + 2021}) = 2021(b – \sqrt{b^2 + 2021})
⇔ – 2021(a + \sqrt{a² + 2021}) = 2021(b – \sqrt{b^2 + 2021})
⇔ a + \sqrt{a² + 2021} = \sqrt{b^2 + 2021} – b (2)
Cộng vế (1) với vế (2), ta có:
b + \sqrt{b² + 2021} + a + \sqrt{a² + 2021} = \sqrt{a^2 + 2021} – a + \sqrt{b^2 + 2021} – b
⇔ a + b + a + b = – \sqrt{b² + 2021} – \sqrt{a² + 2021} + \sqrt{a^2 + 2021} + \sqrt{b^2 + 2021}
⇔ 2(a + b) = 0
⇔ a + b = 0