Giải thích các bước giải:
*
\[\begin{array}{l}
\left( {4x - 1} \right)\sqrt {{x^2} + 1} = 2{x^2} + 2x + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x > \frac{1}{4}\\
{\left( {4x - 1} \right)^2}\left( {{x^2} + 1} \right) = \left( {2{x^2} + 2x + 1} (1) \right)
\end{array} \right.
\end{array}\]
\[\begin{array}{l}
\left( 1 \right) \Leftrightarrow \left( {16{x^2} - 8x + 1} \right)\left( {{x^2} + 1} \right) = 4{x^4} + 4{x^2} + 1 + 8{x^3} + 4x + 4{x^2}\\
\Leftrightarrow 16{x^4} + 16{x^2} - 8{x^3} - 8x + {x^2} + 1 = 4{x^4} + 8{x^3} + 8{x^2} + 4x + 1\\
\Leftrightarrow 12{x^4} - 16{x^3} + 9{x^2} - 12x = 0\\
\Leftrightarrow x\left( {3x - 4} \right)\left( {4{x^2} + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{4}{3}
\end{array} \right.\\
x > \frac{1}{4} \Rightarrow x = \frac{4}{3}
\end{array}\]
*
\[\begin{array}{l}
3\left( {{x^2} + 2x + 2} \right) = 10\sqrt {{x^3} + 2{x^2} + 2x + 1} \\
\Leftrightarrow 3\left( {{x^2} + x + 1} \right) + 3\left( {x + 1} \right) = 10\sqrt {\left( {{x^3} + {x^2} + x} \right) + \left( {{x^2} + x + 1} \right)} \\
\Leftrightarrow 3\left( {{x^2} + x + 1} \right) + 3\left( {x + 1} \right) = 10\sqrt {\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)} \\
\Leftrightarrow 3\left( {{x^2} + x + 1} \right) + 3\left( {x + 1} \right) - 10\sqrt {\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)} \\
\Leftrightarrow 3{a^2} + 3{b^2} - 10ab = 0\\
\Leftrightarrow \left( {3a - b} \right)\left( {a - 3b} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3a = b\\
a = 3b
\end{array} \right.
\end{array}\]
