Cho a,b thỏa mãn: `(a+ √(a ²+2021))(b+ √(b ²+2021)) =2021` Hãy tính `a+b`.

Phương pháp làm chủ yếu là nhân liên hợp

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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`

Đáp án:

Giải thích các bước giải:

`(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`