Giải bài 74 trang 127 SGK giải tích nâng cao 12

a) Điều kiện: $x<1$.

${{\log }_{2}}\left( 3-x \right)+{{\log }_{2}}\left( 1-x \right)=3 \\ \Leftrightarrow \left( 3-x \right)\left( 1-x \right)=8 \\ \Leftrightarrow {{x}^{2}}-4x-5=0 \\ \Leftrightarrow \left[ \begin{aligned} & x=-1 \\ & x=5\,\left( \text{loại} \right) \\ \end{aligned} \right.$

b) Điều kiện: $x<3$.

${{\log }_{2}}\left( 9-{{2}^{x}} \right)={{10}^{\log \left( 3-x \right)}} \\ \Leftrightarrow {{\log }_{2}}\left( 9-{{2}^{x}} \right)=3-x \\ \Leftrightarrow 9-{{2}^{x}}={{2}^{3-x}} \\ \Leftrightarrow 9-{{2}^{x}}=\dfrac{8}{{{2}^{x}}} \\ \Leftrightarrow {{2}^{2x}}-{{9.2}^{x}}+8=0 \\ \Leftrightarrow \left[ \begin{aligned} & {{2}^{x}}=1 \\ & {{2}^{x}}=8 \\ \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} & x=0 \\ & x=3\,\left( \text{loại} \right) \\ \end{aligned} \right.$

c) Điều kiện: $x>1$.

${{7}^{\log x}}-{{5}^{\log x+1}}={{3.5}^{\log x-1}}-{{13.7}^{\log x-1}} \\ {{7}^{\log x}}+\dfrac{13}{7}{{.7}^{\log x}}={{5.5}^{\log x}}+\dfrac{3}{5}{{.5}^{\log x}} \\ \Leftrightarrow \dfrac{20}{7}{{.7}^{\log x}}=\dfrac{28}{5}{{.5}^{\log x}} \\ \Leftrightarrow {{\left( \dfrac{7}{5} \right)}^{\log x}}=\dfrac{49}{25}={{\left( \dfrac{7}{5} \right)}^{2}} \\ \Leftrightarrow \log x=2 \\ \Leftrightarrow x={{10}^{2}}=100$

$d)\, {{6}^{x}}+{{6}^{x+1}}={{2}^{x}}+{{2}^{x+1}}+{{2}^{x+2}} \\ \Leftrightarrow {{7.6}^{x}}={{7.2}^{x}} \\ \Leftrightarrow {{2}^{x}}\left( {{3}^{x}}-1 \right)=0 \\ \Leftrightarrow {{3}^{x}}-1=0\,\left( {{2}^{x}}\ne 0 \right) \\ \Leftrightarrow x=0$