Đáp án:

\(\begin{array}{l} a)\,\,\,x = {\log _2}3.\\ b)\,\,\left[ \begin{array}{l} m = - 1\\ m > 0\\ - 1 0 \end{array} \right.\\ d)\,\, - 1

Giải thích các bước giải: \[\begin{array}{l} {4^x} - {2^{x + 1}} = m\,\,\,\\ \Leftrightarrow {2^{2x}} - {2.2^x} - m = 0\,\left( * \right)\\ \,Dat\,\,\,t = {2^x}\,\,\,\left( {t > 0} \right)\\ \Rightarrow \left( * \right) \Leftrightarrow {t^2} - 2t - m = 0\,\,\,\,\left( 1 \right)\\ a)\,\,Giai\,\,pt\,\,vs\,\,\,m = 3.\\ Voi\,\,\,m = 3\,\,\, \Rightarrow \left( * \right) \Leftrightarrow {2^2}^x - {2.2^x} - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l} {2^x} = 3\,\,\,\,\left( {tm} \right)\\ {2^x} = - 1\,\,\,\left( {ktm} \right) \end{array} \right. \Leftrightarrow x = {\log _2}3.\\ b)\,\,\left( * \right)\,\,\,co\,\,\,nghiem\,\,\, \Leftrightarrow \left( 1 \right)\,\,\,\,co\,\,\,\,nghiem\,\,\,t > 0\\ \Leftrightarrow \left[ \begin{array}{l} \left( 1 \right)\,\,\,co\,\,\,nghiem\,\,kep\,\,duong\\ \left( 1 \right)\,\,\,co\,\,\,nghiem\,\,\,trai\,\,\,dau\\ \left( 1 \right)\,\,\,co\,\,2\,\,nghiem\,\,\,duong\,\,\,phan\,\,biet \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \Delta ' = 0\\ - \frac{b}{a} > 0\\ \frac{c}{a} > 0 \end{array} \right.\\ ac 0\\ - \frac{b}{a} > 0\\ \frac{c}{a} > 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 1 + m = 0\\ 2 > 0\\ - m > 0 \end{array} \right.\\ - m 0\\ 2 > 0\\ - m > 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} m = - 1\\ m 0\\ \left\{ \begin{array}{l} m > - 1\\ m 0\\ - 1 0\\ \frac{c}{a} > 0 \end{array} \right.\\ ac 0\\ - m > 0 \end{array} \right.\\ - m 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = - 1\\ m > 0 \end{array} \right..\\ d)\,\,\left( * \right)\,\,\,co\,\,\,2\,\,nghiem\,\,\,pb \Leftrightarrow \left( 1 \right)\,\,\,co\,\,2\,\,\,nghiem\,\,\,duong\,\,\,phan\,\,biet\\ \Leftrightarrow \left\{ \begin{array}{l} \Delta ' > 0\\ - \frac{b}{a} > 0\\ \frac{c}{a} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 1 + m > 0\\ 2 > 0\\ - m > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m > - 1\\ m