Đáp án:
d2) x=1
Giải thích các bước giải:
\(\begin{array}{l}
d1)\left| {2x + 6} \right| = x + 3\\
TH1:2x + 6 \ge 0 \to x \ge - 3\\
Pt \to 2x + 6 = x + 3\\
\to x = - 3\left( {TM} \right)\\
TH2:2x + 6 < 0 \to x < - 3\\
Pt \to 2x + 6 = - x - 3\\
\to 3x = - 9\\
\to x = - 3\left( {KTM} \right)\\
KL:x = - 3\\
d2)\left| {x - 3} \right| = 5 - 3x\\
TH1:x - 3 \ge 0\\
\to x \ge 3\\
Pt \to x - 3 = 5 - 3x\\
\to 4x = 8\\
\to x = 2\left( {KTM} \right)\\
TH2:x - 3 < 0 \to x < 3\\
Pt \to x - 3 = - 5 + 3x\\
\to 2x = 2\\
\to x = 1\left( {TM} \right)\\
KL:x = 1
\end{array}\)
\(\begin{array}{l}
a)7x - 14 = 0\\
\to x = 2\\
5x + 7 = 10\\
\to 5x = 3\\
\to x = \dfrac{3}{5}\\
6x - 18 = 0\\
\to x = 3\\
b)\dfrac{{2x + 3x - 1 - 2.6 + 2x.4}}{{12}} = 0\\
\to 13x - 13 = 0\\
\to x = 1\\
\dfrac{{2x + 2x - 1 - 10}}{{10}} = - \dfrac{2}{3}x\\
\to \dfrac{{4x - 11}}{{10}} = - \dfrac{2}{3}x\\
\to 12x - 33 = - 20x\\
\to 32x = 33\\
\to x = \dfrac{{33}}{{32}}\\
c)DK:x \ne \pm 2\\
\dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2} - 4{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to {x^2} + 4x + 4 - {x^2} + 4x - 4 - 4{x^2} = 0\\
\to 8x - 4{x^2} = 0\\
\to 4x\left( {2 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2\left( l \right)
\end{array} \right.\\
DK:x \ne \pm 3\\
\dfrac{{{{\left( {x + 3} \right)}^2} - {{\left( {x - 3} \right)}^2} - 18}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\\
\to {x^2} + 6x + 9 - {x^2} + 6x - 9 - 18 = 0\\
\to 12x - 18 = 0\\
\to x = \dfrac{3}{2}
\end{array}\)
