1,(2x-1)^2+(4x^2-1)^2=0 2,x^3+3x^2+3x=7

Bởi

1,(2x-1)^2+(4x^2-1)^2=0
2,x^3+3x^2+3x=7

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  2. Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    1,\\
    {\left( {2x – 1} \right)^2} + {\left( {4{x^2} – 1} \right)^2} = 0\\
    \left. \begin{array}{l}
    {\left( {2x – 1} \right)^2} \ge 0,\,\,\,\,\forall x\\
    {\left( {4{x^2} – 1} \right)^2} \ge 0,\,\,\,\forall x
    \end{array} \right\} \Rightarrow {\left( {2x – 1} \right)^2} + {\left( {4{x^2} – 1} \right)^2} \ge 0,\,\,\,\forall x\\
     \Rightarrow \left\{ \begin{array}{l}
    {\left( {2x – 1} \right)^2} = 0\\
    {\left( {4{x^2} – 1} \right)^2} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    2x – 1 = 0\\
    4{x^2} – 1 = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x = \frac{1}{2}\\
    {x^2} = \frac{1}{4}
    \end{array} \right. \Leftrightarrow x = \frac{1}{2}\\
    2,\\
    {x^3} + 3{x^2} + 3x = 7\\
     \Leftrightarrow {x^3} + 3{x^2} + 3x – 7 = 0\\
     \Leftrightarrow \left( {{x^3} – {x^2}} \right) + \left( {4{x^2} – 4x} \right) + \left( {7x – 7} \right) = 0\\
     \Leftrightarrow {x^2}\left( {x – 1} \right) + 4x\left( {x – 1} \right) + 7\left( {x – 1} \right) = 0\\
     \Leftrightarrow \left( {x – 1} \right)\left( {{x^2} + 4x + 7} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 1 = 0\\
    {x^2} + 4x + 7 = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = 1\\
    \left( {{x^2} + 4x + 4} \right) + 3 = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = 1\\
    {\left( {x + 2} \right)^2} + 3 = 0\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)
    \end{array} \right.\\
     \Leftrightarrow x = 1
    \end{array}\)

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