Cho ( x + $\sqrt[]{a+x^2}$ )(y + $\sqrt[]{a+y^2}$ ) =a Tính $x^{2a+1}$ + $y^{2a+1}$ với a nguyên dương Giúp em vớiiiiiiiiii

Đáp án:

$x^{2a +1} + y^{2a +1} =0$

Giải thích các bước giải:

$(x + \sqrt{a + x^2})(y +\sqrt{a + y^2})=a \qquad (*)$

$(*)\Leftrightarrow (x – \sqrt{a + x^2})(x + \sqrt{a + x^2})(y +\sqrt{a + y^2})=a(x – \sqrt{a + x^2})$

$\Leftrightarrow [x^2 – (a + x^2)](y +\sqrt{a + y^2})=a(x + \sqrt{a – x^2})$

$\Leftrightarrow -a(y +\sqrt{a + y^2})=a(x -\sqrt{a + x^2})$

$\Leftrightarrow y + \sqrt{x + y^2}=\sqrt{a + x^2} – x\qquad (1)$

$(*)\Leftrightarrow (x + \sqrt{a + x^2})(y +\sqrt{a + y^2})(y -\sqrt{a + y^2}) = a(y -\sqrt{a – y^2})$

$\Leftrightarrow (x + \sqrt{a + x^2})[y^2 – (a+y^2)]=a(y -\sqrt{a – y^2})$

$\Leftrightarrow (x + \sqrt{a + x^2}).(-a) = a(y -\sqrt{a – y^2})$

$\Leftrightarrow x + \sqrt{a + x^2} = \sqrt{a + y^2} – y \qquad (2)$

Lấy $(1)+(2)$ ta được:

$y + \sqrt{x + y^2} + x + \sqrt{a + x^2} = \sqrt{a + x^2} – x + \sqrt{a + y^2} – y$

$\Leftrightarrow 2(x + y) = 0$

$\Leftrightarrow x + y = 0$

$\Leftrightarrow x = – y$

$\Rightarrow x^{2a +1} + y^{2a +1} = (-y)^{2a +1} + y^{2a + 1} = 0$