Đáp án:
\(\begin{array}{l}
30)\quad S = \{16\}\\
31)\quad S = \{-3;-2;1\}\\
32)\quad S = \{0;1\}\\
33)\quad S = \{-1;2\}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
30)\quad 8.3^{\sqrt x + \sqrt[4]{x}} + 9^{1 + \sqrt[4]{x}} = 9^{\sqrt x}\quad (ĐK: x\geqslant 0)\\
\Leftrightarrow 8.3^{\sqrt x}.3^{\sqrt[4]{x}} + 9.3^{2\sqrt[4]{x}} - 3^{2\sqrt x} = 0\\
\Leftrightarrow \left(3^{\sqrt x} + 3^{\sqrt[4]{x}}\right)\left(9.3^{\sqrt[4]{x}} - 3^{\sqrt x}\right) = 0\\
\Leftrightarrow \left[\begin{array}{l}3^{\sqrt x} + 3^{\sqrt[4]{x}} = 0\quad (vn)\\9.3^{\sqrt[4]{x}} - 3^{\sqrt x} = 0\end{array}\right.\\
\Leftrightarrow 9.3^{\sqrt[4]{x}} = 3^{\sqrt x}\\
\Leftrightarrow 2 + \sqrt[4]{x} = \sqrt x\\
\Leftrightarrow \sqrt x - \sqrt[4]{x} - 2 = 0\\
\leftrightarrow \left(\sqrt[4]{x} + 1\right)\left(\sqrt[4]{x} - 2\right) = 0\\
\Leftrightarrow \left[\begin{array}{l}\sqrt[4]{x} + 1 = 0\quad (vn)\\\sqrt[4]{x} = 2\end{array}\right.\\
\Leftrightarrow x = 16\\
\text{Vậy}\ S = \{16\}\\
31)\quad 2^{2\sqrt{x+3} - x} - 5.2^{\sqrt{x+3} +1} + 2^{x+4} = 0\quad (ĐK:x\geqslant -3)\\
\Leftrightarrow \left(2^{\sqrt{x+3}} - x\right)^2 - 10.2^{\sqrt{x+3} - x} + 16 = 0\\
\Leftrightarrow \left(2^{\sqrt{x+3} - x} - 2\right)\left(2^{\sqrt{x+3} - x} -8\right) = 0\\
\Leftrightarrow \left[\begin{array}{l}2^{\sqrt{x+3} - x} = 2\\2^{\sqrt{x+3} - x} = 8\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\sqrt{x+3} -x = 1\\\sqrt{x+3} - x = 3\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\sqrt{x+3}= x+1\\\sqrt{x+3}=x+ 3\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\begin{cases}x + 1 \geqslant 0\\x + 3 = (x+1)^2\end{cases}\\x+3 = (x+3)^2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\begin{cases}x \geqslant -1\\x^2+x - 2 =0\end{cases}\\(x+3)(x+2) =0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x= 1\\x = -2\\x = -3\end{array}\right.\\
\text{Vậy}\ S = \{-3;-2;1\}\\
32)\quad 4^{2x^2} - 2.4^{x^2 + x} + 4^{2x} = 0\\
\Leftrightarrow \left(4^{x^2}\right)^2 - 2.4^{x^2}.4^x + \left(4^x\right)^x = 0\\
\Leftrightarrow \left(4^{x^2} - 4^x\right)^2 = 0\\
\Leftrightarrow 4^{x^2} = 4^x\\
\Leftrightarrow x^2 = x\\
\Leftrightarrow \left[\begin{array}{l}x = 0\\x = 1\end{array}\right.\\
\text{Vậy}\ S = \{0;1\}\\
33)\quad 2^{2x^2 + 1} - 9.2^{x^2 + x} + 2^{2x + 2} = 0\\
\Leftrightarrow 2.\left(2^{x^2}\right)^2 - 9.2^{x^2}.2^x + 4.\left(2^x\right)^2 = 0\\
\Leftrightarrow \left(2.2^{x^2} - 2^x\right)\left(2^{x^2} - 4.2^{x}\right) = 0\\
\Leftrightarrow \left[\begin{array}{l}2.2^{x^2} =2^x\\2^{x^2} = 4.2^x\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x^2 + 1 = x\\x^2 = x + 2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x^2 - x + 1 =0\quad (vn)\\x^2- x - 2 = 0\end{array}\right.\\
\Leftrightarrow (x+1)(x-2) = 0\\
\Leftrightarrow \left[\begin{array}{l}x= - 1\\x = 2\end{array}\right.\\
\text{Vậy}\ S = \{-1;2\}
\end{array}\)
