Tìm x a)4-11x+7x^2=x-1(x≠1) e)x^3-x=2x^2-2(x≠±1)

Đáp án:

b. x=2

Giải thích các bước giải:

\(\begin{array}{l}
a.4 – 11x + 7{x^2} = x – 1\\
 \to 7{x^2} – 12x + 5 = 0\\
 \to 7{x^2} – 7x – 5x + 5 = 0\\
 \to 7x\left( {x – 1} \right) – 5\left( {x – 1} \right) = 0\\
 \to \left( {x – 1} \right)\left( {7x – 5} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{5}{7}
\end{array} \right.\\
Do:x \ne 1\\
 \to x = \dfrac{5}{7}\\
b.{x^3} – x = 2{x^2} – 2\\
 \to {x^3} – 2{x^2} – x + 2 = 0\\
 \to {x^3} – {x^2} – {x^2} + x – 2x + 2 = 0\\
 \to {x^2}\left( {x – 1} \right) – x\left( {x – 1} \right) – 2\left( {x – 1} \right) = 0\\
 \to \left( {x – 1} \right)\left( {{x^2} – x – 2} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 1\\
\left( {x – 2} \right)\left( {x + 1} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  – 1\\
x = 2
\end{array} \right.\\
Do:x \ne  \pm 1\\
 \to x = 2
\end{array}\)