Question

Đã trả lời bởi chuyên gia

Answer 1

áp án B

Ta có: $\log_{3} {[(x + 1) (y + 1)]}^{y + 1} = 9 - (x - 1) (y + 1) \Leftrightarrow (y + 1) \log_{3} [(x + 1) (y + 1)] + (x - 1) (y + 1) = 9$

$\Leftrightarrow (y + 1) \log_{3} [(c + 1) (y + 1)] + (x + 1) (y + 1) - 2 y = 11$

$\Leftrightarrow (y + 1) (\log_{3} [(c + 1) (y + 1)] - 2) = 9 - (x + 1) (y + 1) (*)$

Nếu $(x + 1) (y + 1) > 9 \Rightarrow V T (*) > 0; V P (*) < 0$

Ngược lại nếu $(x + 1) (y + 1) < 9 \Rightarrow V T (*) < 0; V P (*) > 0$

Do đó $(*) \Leftrightarrow (x + 1) (y + 1) = 9 \Leftrightarrow x y + x + y = 8$

Khi đó $P = {(x + y)}^{3} - 3 x y (x + y) - 57 (x + y) = {(x + y)}^{3} - 3 (8 - x - y) (x + y) - 57 (x + y)$

Đặt $t = (x + y) \geq 2 \Rightarrow f (t) = t^{3} - 3 (8 - t) t - 57 t = t^{3} + 3 t^{2} - 81 t$

$\Rightarrow f' (t) = 3 t^{2} + 6 t - 81 = 0 \Rightarrow t = - 1 + 2 \sqrt{7} \Rightarrow P_{\min} = f (- 1 + 2 \sqrt{7}) = 83 - 112 \sqrt{7} \Rightarrow a + b = - 29$

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Answer 2