a) |2x-1|=|2x-5|
b) |7-x | -|2x-3|=0
c) |x-4|+|x²-5x+4|=0
d) |x²-x-2 phần x+1| – |x|=0
a) |2x-1|=|2x-5| b) |7-x | -|2x-3|=0 c) |x-4|+|x²-5x+4|=0 d) |x²-x-2 phần x+1| – |x|=0
By Adalynn
By Adalynn
a) |2x-1|=|2x-5|
b) |7-x | -|2x-3|=0
c) |x-4|+|x²-5x+4|=0
d) |x²-x-2 phần x+1| – |x|=0
Đáp án:
c) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {2x – 1} \right| = \left| {2x – 5} \right|\\
\to \left[ \begin{array}{l}
2x – 1 = 2x – 5\left( l \right)\\
2x – 1 = – 2x + 5
\end{array} \right.\\
\to 4x = 6\\
\to x = \dfrac{3}{2}\\
b)\left| {7 – x} \right| = \left| {2x – 3} \right|\\
\to \left[ \begin{array}{l}
7 – x = 2x – 3\left( {DK:x \le 7} \right)\\
– 7 + x = 2x – 3\left( {DK:x > 7} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 10\\
x = – 4\left( l \right)
\end{array} \right.\\
\to x = \dfrac{{10}}{3}\\
c)\left| {x – 4} \right| + \left| {{x^2} – 5x + 4} \right| = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x – 4 = 0\\
{x^2} – 5x + 4 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
\left( {x – 1} \right)\left( {x – 4} \right) = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
x = 1\\
x = 4
\end{array} \right.\\
KL:x = 4\\
d)\left| {\dfrac{{{x^2} – x – 2}}{{x + 1}}} \right| – \left| x \right| = 0\\
\to \left| {\dfrac{{{x^2} – x – 2}}{{x + 1}}} \right| = \left| x \right|\\
\to \left[ \begin{array}{l}
\dfrac{{{x^2} – x – 2}}{{x + 1}} = x\left( {DK:x \ge 0} \right)\\
\dfrac{{{x^2} – x – 2}}{{x + 1}} = – x\left( {DK:x < 0;x \ne – 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} – x – 2 = {x^2} + x\\
{x^2} – x – 2 = – {x^2} – x
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = – 2\\
2{x^2} = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 1\left( l \right)\\
x = 1\left( l \right)
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)