$(x+3)^{4}$ – ($x^{2}$+x-6)^2 = 2$(x-2)^{4}$ 01/08/2021 Bởi Piper $(x+3)^{4}$ – ($x^{2}$+x-6)^2 = 2$(x-2)^{4}$
Đáp số: \[\left[ \begin{array}{l}x = 7 + 5\sqrt 2 \\x = 7 – 5\sqrt 2 \end{array} \right.\] Giải thích các bước giải: Ta có: \(\begin{array}{l}{\left( {x + 3} \right)^4} – {\left( {{x^2} + x – 6} \right)^2} = 2{\left( {x – 2} \right)^4}\\ \Leftrightarrow {\left( {x + 3} \right)^4} – {\left[ {\left( {{x^2} + 3x} \right) – \left( {2x + 6} \right)} \right]^2} – 2{\left( {x – 2} \right)^4} = 0\\ \Leftrightarrow {\left( {x + 3} \right)^4} – {\left[ {x.\left( {x + 3} \right) – 2\left( {x + 3} \right)} \right]^2} – 2{\left( {x – 2} \right)^4} = 0\\ \Leftrightarrow {\left( {x + 3} \right)^4} – {\left[ {\left( {x + 3} \right)\left( {x – 2} \right)} \right]^2} – 2.{\left( {x – 2} \right)^4} = 0\\ \Leftrightarrow {\left( {x + 3} \right)^4} – {\left( {x + 3} \right)^2}{\left( {x – 2} \right)^2} – 2{\left( {x – 2} \right)^4} = 0\\ \Leftrightarrow \left[ {{{\left( {x + 3} \right)}^4} + {{\left( {x + 3} \right)}^2}{{\left( {x – 2} \right)}^2}} \right] – \left[ {2{{\left( {x + 3} \right)}^2}{{\left( {x – 2} \right)}^2} + 2{{\left( {x – 2} \right)}^4}} \right] = 0\\ \Leftrightarrow {\left( {x + 3} \right)^2}\left[ {{{\left( {x + 3} \right)}^2} + {{\left( {x – 2} \right)}^2}} \right] – 2{\left( {x – 2} \right)^2}.\left[ {{{\left( {x + 3} \right)}^2} + {{\left( {x – 2} \right)}^2}} \right] = 0\\ \Leftrightarrow \left[ {{{\left( {x + 3} \right)}^2} + {{\left( {x – 2} \right)}^2}} \right]\left[ {{{\left( {x + 3} \right)}^2} – 2.{{\left( {x – 2} \right)}^2}} \right] = 0\\ \Leftrightarrow \left[ {{x^2} + 6x + 9 + {x^2} – 4x + 4} \right].\left[ {\left( {{x^2} + 6x + 9} \right) – 2.\left( {{x^2} – 4x + 4} \right)} \right] = 0\\ \Leftrightarrow \left( {2{x^2} + 2x + 13} \right).\left( { – {x^2} + 14x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2{x^2} + 2x + 13 = 0\\ – {x^2} + 14x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}{x^2} + \left( {{x^2} + 2x + 1} \right) + 12 = 0\\{x^2} – 14x – 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}{x^2} + {\left( {x + 1} \right)^2} + 12 = 0\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\{x^2} – 2.7.x + 49 = 50\end{array} \right.\\ \Leftrightarrow {\left( {x – 7} \right)^2} = 50\\ \Leftrightarrow \left[ \begin{array}{l}x – 7 = 5\sqrt 2 \\x – 7 = – 5\sqrt 2 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 7 + 5\sqrt 2 \\x = 7 – 5\sqrt 2 \end{array} \right.\end{array}\) Bình luận
Đáp án: Đặt (x+3)^2=a,(x-2)^2=b đk a,b>=0 ta có phương trình a^2-ab=2b^2 (=) a=2b hoặc a=-b tự giải tiếp Bình luận
Đáp số:
\[\left[ \begin{array}{l}
x = 7 + 5\sqrt 2 \\
x = 7 – 5\sqrt 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {x + 3} \right)^4} – {\left( {{x^2} + x – 6} \right)^2} = 2{\left( {x – 2} \right)^4}\\
\Leftrightarrow {\left( {x + 3} \right)^4} – {\left[ {\left( {{x^2} + 3x} \right) – \left( {2x + 6} \right)} \right]^2} – 2{\left( {x – 2} \right)^4} = 0\\
\Leftrightarrow {\left( {x + 3} \right)^4} – {\left[ {x.\left( {x + 3} \right) – 2\left( {x + 3} \right)} \right]^2} – 2{\left( {x – 2} \right)^4} = 0\\
\Leftrightarrow {\left( {x + 3} \right)^4} – {\left[ {\left( {x + 3} \right)\left( {x – 2} \right)} \right]^2} – 2.{\left( {x – 2} \right)^4} = 0\\
\Leftrightarrow {\left( {x + 3} \right)^4} – {\left( {x + 3} \right)^2}{\left( {x – 2} \right)^2} – 2{\left( {x – 2} \right)^4} = 0\\
\Leftrightarrow \left[ {{{\left( {x + 3} \right)}^4} + {{\left( {x + 3} \right)}^2}{{\left( {x – 2} \right)}^2}} \right] – \left[ {2{{\left( {x + 3} \right)}^2}{{\left( {x – 2} \right)}^2} + 2{{\left( {x – 2} \right)}^4}} \right] = 0\\
\Leftrightarrow {\left( {x + 3} \right)^2}\left[ {{{\left( {x + 3} \right)}^2} + {{\left( {x – 2} \right)}^2}} \right] – 2{\left( {x – 2} \right)^2}.\left[ {{{\left( {x + 3} \right)}^2} + {{\left( {x – 2} \right)}^2}} \right] = 0\\
\Leftrightarrow \left[ {{{\left( {x + 3} \right)}^2} + {{\left( {x – 2} \right)}^2}} \right]\left[ {{{\left( {x + 3} \right)}^2} – 2.{{\left( {x – 2} \right)}^2}} \right] = 0\\
\Leftrightarrow \left[ {{x^2} + 6x + 9 + {x^2} – 4x + 4} \right].\left[ {\left( {{x^2} + 6x + 9} \right) – 2.\left( {{x^2} – 4x + 4} \right)} \right] = 0\\
\Leftrightarrow \left( {2{x^2} + 2x + 13} \right).\left( { – {x^2} + 14x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} + 2x + 13 = 0\\
– {x^2} + 14x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + \left( {{x^2} + 2x + 1} \right) + 12 = 0\\
{x^2} – 14x – 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + {\left( {x + 1} \right)^2} + 12 = 0\,\,\,\,\,\,\,\,\,\left( {vn} \right)\\
{x^2} – 2.7.x + 49 = 50
\end{array} \right.\\
\Leftrightarrow {\left( {x – 7} \right)^2} = 50\\
\Leftrightarrow \left[ \begin{array}{l}
x – 7 = 5\sqrt 2 \\
x – 7 = – 5\sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7 + 5\sqrt 2 \\
x = 7 – 5\sqrt 2
\end{array} \right.
\end{array}\)
Đáp án:
Đặt (x+3)^2=a,(x-2)^2=b đk a,b>=0
ta có phương trình a^2-ab=2b^2
(=) a=2b hoặc a=-b
tự giải tiếp