125^x+50^x-2^(3x+1)=0

Đáp án: x=0

 Giải thích các bước giải:

`125^x+50^x-2^(3x+1)=0`

`⇔125^x+50^x=2^(3x+1)`

`⇔5^(3x)+(2.5^2)^x=2.2^(3x)`

`⇔(5/2)^(3x)+(5/2)^(2x)=2`

`⇔[(5/2)^x]^3+[(5/2)^x]^2=2`

Đặt `t=(5/2)^x` , ta có:

`t^3+t^2=2`

`⇔t^3+t^2-2=0`

`⇔t=1`

`⇒(5/2)^x=1 ⇔ x=0`

Đáp án:

$S=\{0\}$

Giải thích các bước giải:

$\begin{array}{l}125^{\displaystyle{x}}+50^{\displaystyle{x}}-2^{\displaystyle{3x+1}} =0\\ \Leftrightarrow 5^{\displaystyle{3x}} + (5^2.2)^{\displaystyle{x}} - 2.2^{\displaystyle{3x}} = 0\\ \Leftrightarrow 5^{\displaystyle{3x}} + 5^{\displaystyle{2x}}.2^{\displaystyle{x}} - 2.2^{\displaystyle{3x}} = 0\\ \Leftrightarrow 5^{\displaystyle{3x}} -2^{\displaystyle{3x}}+ 5^{\displaystyle{2x}}.2^{\displaystyle{x}} - 2^{\displaystyle{3x}} = 0\\ \Leftrightarrow (5^{\displaystyle{x}} -2^{\displaystyle{x}})(5^{\displaystyle{2x}} + 5^{\displaystyle{x}}.2^{\displaystyle{x}}+2^{\displaystyle{2x}}) +2^{\displaystyle{x}}(5^{\displaystyle{2x}} -2^{\displaystyle{2x}}) = 0\\ \Leftrightarrow (5^{\displaystyle{x}} -2^{\displaystyle{x}})(5^{\displaystyle{2x}} + 5^{\displaystyle{x}}.2^{\displaystyle{x}}+2^{\displaystyle{2x}}) +2^{\displaystyle{x}}(5^{\displaystyle{x}} -2^{\displaystyle{x}})(5^{\displaystyle{x}}+2^{\displaystyle{x}}) = 0\\ \Leftrightarrow (5^{\displaystyle{x}}-2^{\displaystyle{x}})(5^{\displaystyle{2x}} +5^{\displaystyle{x}}.2^{\displaystyle{x}}+2^{\displaystyle{2x}}+2^{\displaystyle{x}}.5^{\displaystyle{x}}-2^{\displaystyle{2x}}) =0\\ \Leftrightarrow (5^{\displaystyle{x}}-2^{\displaystyle{x}})(5^{\displaystyle{2x}}+2.5^{\displaystyle{x}}.2^{\displaystyle{x}})=0\\ \Leftrightarrow 5^{\displaystyle{x}}(5^{\displaystyle{x}}-2^{\displaystyle{x}})(5^{\displaystyle{x}}+2.2^{\displaystyle{x}}) =0\\ \Leftrightarrow 5^{\displaystyle{x}}-2^{\displaystyle{x}}=0\\ \Leftrightarrow 5^{\displaystyle{x}}=2^{\displaystyle{x}}\\ \Leftrightarrow \left(\dfrac52\right)^{\displaystyle{x}}=1\\ \Leftrightarrow x = 0\\ \text{Vậy phương trình có tập nghiệm}\,\,S=\{0\} \end{array}$