Tìm x a, 2 . ( x + 1 ) – ( x + 3 ) . ( x + 3 ) – ( x – 4 ) =0 b, ( x – 5 ) – x. ( x – 4 ) = 9 c, ( x – 5 ) + ( x – 4 ) . ( 1 – x ) = 0

Đáp án:

c. x=3

Giải thích các bước giải:

\(\begin{array}{l}
a.2\left( {x + 1} \right) – \left( {x + 3} \right)\left( {x + 3} \right) – \left( {x – 4} \right) = 0\\
 \to 2x + 2 – {x^2} – 6x – 9 – x + 4 = 0\\
 \to  – {x^2} – 5x – 3 = 0\\
 \to  – \left( {{x^2} + 5x + 3} \right) = 0\\
 \to {x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} – \dfrac{{13}}{4} = 0\\
 \to {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{13}}{4}\\
 \to \left[ \begin{array}{l}
x + \dfrac{5}{2} = \dfrac{{\sqrt {13} }}{2}\\
x + \dfrac{5}{2} =  – \dfrac{{\sqrt {13} }}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{ – 5 + \sqrt {13} }}{2}\\
x = \dfrac{{ – 5 – \sqrt {13} }}{2}
\end{array} \right.\\
b.\left( {x – 5} \right) – x\left( {x – 4} \right) = 9\\
 \to x – 5 – {x^2} + 4x – 9 = 0\\
 \to {x^2} – 5x + 14 = 0\\
 \to {x^2} – 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} + \dfrac{{31}}{4} = 0\\
 \to {\left( {x – \dfrac{5}{2}} \right)^2} + \dfrac{{31}}{4} = 0\\
Do:{\left( {x – \dfrac{5}{2}} \right)^2} + \dfrac{{31}}{4} > 0\\
 \to x \in \emptyset \\
c.x – 5 + \left( {x – 4} \right)\left( {1 – x} \right) = 0\\
 \to x – 5 + x – {x^2} – 4 + 4x = 0\\
 \to {x^2} – 6x + 9 = 0\\
 \to {\left( {x – 3} \right)^2} = 0\\
 \to x – 3 = 0\\
 \to x = 3
\end{array}\)