Đáp án:
$6) \left[\begin{array}{l} x=\dfrac{5}{3}\\ x=-1\end{array} \right.\\ 7)\left[\begin{array}{l} x=1\\x=2\end{array} \right.\\ 8)\left[\begin{array}{l} x=1\\ x=0\end{array} \right.\\ 9) \left[\begin{array}{l} x=\dfrac{\log_32}{1-\log_32}\\ x=0\end{array} \right.$
Giải thích các bước giải:
$6)\\ 2^{3x^2-2x}=32\\ \Leftrightarrow 2^{3x^2-2x}=2^5\\ \Leftrightarrow 3x^2-2x=5\\ \Leftrightarrow 3x^2-2x-5=0\\ \Leftrightarrow 3x^2+3x-5x-5=0\\ \Leftrightarrow 3x(x+1)-5(x+1)=0\\ \Leftrightarrow (3x-5)(x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{5}{3}\\ x=-1\end{array} \right.\\ 7)\\ 3^{x^2-3x}=\dfrac{1}{9}\\ \Leftrightarrow 3^{x^2-3x}=\dfrac{1}{3^2}\\ \Leftrightarrow 3^{x^2-3x}=3^{-2}\\ \Leftrightarrow x^2-3x=-2\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow x(x-1)-2(x-1)=0\\ \Leftrightarrow (x-2)(x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=1\\x=2\end{array} \right.\\ 8)\\ 9^x-4.3^x+3=0\\ \Leftrightarrow (3^2)^x-4.3^x+3=0\\ \Leftrightarrow (3^x)^2-4.3^x+3=0\\ \Leftrightarrow (3^x)^2-3^x-3.3^x+3=0\\ \Leftrightarrow 3^x(3^x-1)-3(3^x-1)=0\\ \Leftrightarrow (3^x-3)(3^x-1)\\ \Leftrightarrow \left[\begin{array}{l} 3^x=3\\ 3^x=1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=1\\ x=0\end{array} \right.\\ 9)\\ 9^x-3.6^x+2.4^x=0\\ \Leftrightarrow (3^2)^x-3.(3.2)^x+2.(2^2)^x=0\\ \Leftrightarrow (3^x)^2-3.3^x.2^x+2.(2^x)^2=0\\ \Leftrightarrow (3^x)^2-3^x.2^x+2.(2^x)^2-2.3^x.2^x=0\\ \Leftrightarrow 3^x(3^x-2^x)+2.2^x(2^x-3^x)=0\\ \Leftrightarrow (3^x-2.2^x)(3^x-2^x)=0\\ \Leftrightarrow (3^x-2^{x+1})(3^x-2^x)=0\\ \Leftrightarrow (3^x-2^{x+1})(3^x-2^x)=0\\ \Leftrightarrow \left[\begin{array}{l} 3^x=2^{x+1}\\ 3^x=2^x\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \log_33^x=\log_32^{x+1}\\ \log_33^x=\log_32^x\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=(x+1)\log_32\\ x=x\log_32\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=x\log_32+\log_32\\ x-x\log_32=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x-x\log_32=\log_32\\ x(1-\log_32)=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x(1-\log_32)=\log_32\\ x=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{\log_32}{1-\log_32}\\ x=0\end{array} \right.$
