Đáp án:
b) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{x - 30}}{{10}} + \dfrac{{x - 35}}{{15}} + \dfrac{{x - 47}}{{27}} = - 3\\
\to \dfrac{{27\left( {x - 30} \right) + 18\left( {x - 35} \right) + 10\left( {x - 47} \right)}}{{270}} = 0\\
\to 27x - 810 + 18x - 630 + 10x - 470 = 0\\
\to 55x = 1910\\
\to x = \dfrac{{382}}{{11}}\\
b){x^3} - 4{x^2} + x - 4 = 0\\
\to {x^2}\left( {x - 4} \right) + \left( {x - 4} \right) = 0\\
\to \left( {x - 4} \right)\left( {{x^2} + 1} \right) = 0\\
\to x - 4 = 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\
\to x = 4
\end{array}\)
