Решите уравнение: [tex] 1) 27^{3x-1} = \frac{1}{9} ^{x} 2) 5^{3x} +3* 5^{3x-2} =140 3)

Ответы 1

$1)\\\\ 27^{3x-1}=\frac{1}{9}^{x}\\\\ (3^3)^{3x-1}=(3^{-2})^{x}\\\\ 3^{9x-3}=3^{-2x}\\\\ \Rightarrow 9x-3=-2x\\\\ 9x-3+2x=0\\\\ 11x=3\\\\ \boxed{x=\frac{3}{11}}$ $2)\\\\ 5^{3x}+3 \cdot 5^{3x-2}=140\\\\ 5^{3x-2} \cdot 5^{2}+3 \cdot 5^{3x-2}=140\\\\ 5^{3x-2} (5^{2}+3)=140\\\\ 5^{3x-2} \cdot 28=140\\\\ 5^{3x-2}=140:28\\\\ 5^{3x-2}=5^{1}\\\\ \Rightarrow 3x-2=1\\\\ 3x=1+2\\\\ 3x=3\\\\ x=3:3\\\\ \boxed{x=1}$ $3)\\\\ 4^{x}-2^{x}-12=0\\\\ (2^{2})^{x}-2^{x}=12\\\\ 2^{2x}-2^{x}=12\\\\ 2^{x}(2^{2}-1)=12\\\\ 2^{x} \cdot 3=12\\\\ 2^{x}=12:3\\\\ 2^{x}=4\\\\ 2^{x}=2^{2}\\\\ \Rightarrow \boxed{x=2}$ $5)\\\\ (\frac{3}{5})^{x^{2}-x}=(\frac{5}{3})^{-2}\\\\ (\frac{3}{5})^{x^{2}-x}=[(\frac{3}{5})^{-1}]^{-2}\\\\ (\frac{3}{5})^{x^{2}-x}=(\frac{3}{5})^{2}\\\\ \Rightarrow x^{2}-x=2\\\\ x^{2}-x-2=0\\\\ \Delta=b^{2}-4ac\\\\ a=1;~b=-1;~c=-2\\\\ \Delta=(-1)^{2}-4\cdot1\cdot(-2)\\\\ \Delta=1+8=9\\\\ x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}\\\\x_{1,2}=\frac{-(-1) \pm \sqrt{9}}{2\cdot1}\\\\ x_{1,2}=\frac{1 \pm 3}{2}\\\\ x_{1}=\frac{1+3}{2}=\frac{4}{2}=2\\\\ x_{2}=\frac{1-3}{2}=\frac{-2}{2}=-1\\\\ x \in \{ -1;~2 \}$
- Автор:
  clifford
- 5 лет назад
- 0
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