giải pt:(x+can2)^4+(x+1)^4=33+12can2

Đáp án:

\(\left[ \matrix{ x = 1 \hfill \cr x \approx - 3.41 \hfill \cr} \right.\)

Giải thích các bước giải:

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