Đáp án:
d. \(\left[ \begin{array}{l}
x > 2\\
0 < x < 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{21\left( {4x + 3} \right)}}{{105}} - \dfrac{{15\left( {6x - 2} \right)}}{{105}} = \dfrac{{35\left( {5x + 4} \right)}}{{105}} + \dfrac{{315}}{{105}}\\
\to 84x + 63 - 90x + 30 = 175x + 140 + 315\\
\to 181x = - 362\\
\to x = - 2\\
b.DK:x \ne 8\\
\dfrac{{3.12 + 24\left( {3x - 20} \right) + 3\left( {x - 8} \right)}}{{24\left( {x - 8} \right)}} = \dfrac{{8\left( {13x - 102} \right)}}{{24\left( {x - 8} \right)}}\\
\to 36 + 72x - 480 + 3x - 24 = 104x - 816\\
\to 29x = 348\\
\to x = 12\\
c.\dfrac{{20x - 12 + 10x + 5}}{{20}} < \dfrac{{20 - 30x - 100}}{{20}}\\
\to 30x - 7 < - 30x - 80\\
\to 60x < - 73\\
\to x < - \dfrac{{73}}{{60}}\\
d.DK:x \ne \left\{ {0;2} \right\}\\
\dfrac{{{x^2} + {x^2} - 4 - 2x\left( {x - 2} \right)}}{{x\left( {x - 2} \right)}} > 0\\
\to \dfrac{{2{x^2} - 4 - 2{x^2} + 4x}}{{x\left( {x - 2} \right)}} > 0\\
\to \dfrac{{4x - 4}}{{x\left( {x - 2} \right)}} > 0\\
\to \dfrac{{x - 1}}{{x\left( {x - 2} \right)}} > 0\\
TH1:x - 1 > 0 \to x > 1\\
\to x\left( {x - 2} \right) > 0\\
\to \left[ \begin{array}{l}
x > 2\\
x < 0
\end{array} \right.\\
\to x > 2\\
TH2:x - 1 < 0 \to x < 1\\
\to x\left( {x - 2} \right) < 0\\
\to 0 < x < 2\\
\to 0 < x < 1\\
KL:\left[ \begin{array}{l}
x > 2\\
0 < x < 1
\end{array} \right.
\end{array}\)
