Đáp án:
\(\left[ \matrix{
x = 1 \hfill \cr
x \approx - 3.41 \hfill \cr} \right.\)
Giải thích các bước giải:
$$\eqalign{
& {\left( {x + \sqrt 2 } \right)^4} + {\left( {x + 1} \right)^4} = 33 + 12\sqrt 2 \cr
& \Leftrightarrow {x^4} + 4\sqrt 2 {x^3} + 12{x^2} + 8\sqrt 2 x + 4 + {x^4} + 4{x^3} + 6{x^2} + 4x + 1 = 33 + 12\sqrt 2 \cr
& \Leftrightarrow 2{x^4} + \left( {4\sqrt 2 + 4} \right){x^3} + 18{x^2} + \left( {8\sqrt 2 + 4} \right)x - 28 - 12\sqrt 2 = 0 \cr
& \Leftrightarrow {x^4} + \left( {2\sqrt 2 + 2} \right){x^3} + 9{x^2} + \left( {4\sqrt 2 + 2} \right)x - 14 - 6\sqrt 2 = 0 \cr
& \Leftrightarrow \left( {x - 1} \right)\left[ {{x^3} + \left( {2\sqrt 2 + 3} \right){x^2} + \left( {2\sqrt 2 + 12} \right)x + 6\sqrt 2 + 14} \right] = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 1 \hfill \cr
{x^3} + \left( {2\sqrt 2 + 3} \right){x^2} + \left( {2\sqrt 2 + 12} \right)x + 6\sqrt 2 + 14 = 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = 1 \hfill \cr
x \approx - 3.41 \hfill \cr} \right. \cr} $$
