Giải pt sau $2^x = x^2$

2 câu trả lời

`2^{x} = x^{2}`

`-> ln (x^2) = ln (2^x)`

`-> 2ln |x| = xln (2)`

`-> x^{-1}.ln |x| = 1/(2)ln (2)`

`text{Với x > 0}`

`-> x^{-1}ln (x) = 1/(2)ln (2)`

`-> W (-ln (x).e^{-ln x}) = W (-ln (sqrt{2})`

`-> -ln (x) = W (-ln\ sqrt{2})`

`-> x = e^{-W (-ln\ sqrt{2})}`

`text{Với x < 0}`

`-> (-x)^(-1).ln (-x) = -1/(2)ln (2)`

`-> W (-ln (-x).e^{-ln (-x)} = W (ln\ sqrt{2})`

`-> -ln (-x) = W (ln\ sqrt{2})`

`-> x = -e^{-W (ln\ \sqrt{2})}`

`-> S = {e^{-W (-ln\ sqrt{2})}; -e^{-W (ln\ \sqrt{2})}}`

 

$\begin{array}{l}x^2 = 2^x\quad (*)\\ +)\,\,Với\,\,x=0\\ (*) \Leftrightarrow 0^2 = 2^0 \quad (vô\,\,lí)\\ \Rightarrow \text{x = 0 không là nghiệm của phương trình}\\ +)\,\,Với\,\,x \ne 0\\ (*) \Leftrightarrow \ln(x^2) = \ln(2^x)\\ \Leftrightarrow 2\ln|x| = x\ln2\\ \Leftrightarrow \dfrac{1}{x}\ln|x| = \dfrac{1}{2}\ln2\\ \Leftrightarrow x^{-1}\ln|x| = \dfrac{1}{2}\ln2\\ +)\,\,Với\,\,x>0,\,ta \,\,được:\\ x^{-1}\ln x = \dfrac{1}{2}\ln2\\ \Leftrightarrow e^{\ln x^{-1}}.\ln x = \ln2^{\tfrac{1}{2}}\\ \Leftrightarrow -\ln x.e^{-\ln x} = -\ln\sqrt2\\ \Leftrightarrow W\left(-\ln x.e^{-\ln x}\right) = W\left(-\ln\sqrt2\right)\\ \Leftrightarrow -\ln x = W\left(-\ln\sqrt2\right)\\ \Leftrightarrow \ln x = -W\left(-\ln\sqrt2\right)\\ \Leftrightarrow e^{\ln x} = e^{-W\left(-\ln\sqrt2\right)}\\ \Leftrightarrow x = e^{-W\left(-\ln\sqrt2\right)}\\ \Leftrightarrow \left[\begin{array}{l}x = e^{-W_o\left(-\ln\sqrt2\right)} = 2\\x = e^{-W_{-1}\left(-\ln\sqrt2\right)} = 4\\ \end{array}\right.\\ +)\,\,Với\,\,x < 0, \,ta\,\,được:\\ x^{-1}\ln(-x) = \dfrac{1}{2}\ln2\\ \Leftrightarrow -x^{-1}\ln(-x) = - \dfrac{1}{2}\ln2\\ \Leftrightarrow (-x)^{-1}\ln(-x) = -\ln2^{\tfrac{1}{2}}\\ \Leftrightarrow e^{-\ln(-x)}\ln(-x) = -\ln\sqrt2\\ \Leftrightarrow -\ln(-x).e^{-\ln(-x)} = \ln\sqrt2\\ \Leftrightarrow W\left(-\ln(-x).e^{-\ln(-x)}\right)=W\left(\ln\sqrt2\right)\\ \Leftrightarrow -\ln(-x) = W\left(\ln\sqrt2\right)\\ \Leftrightarrow \ln(-x) = -W\left(\ln\sqrt2\right)\\ \Leftrightarrow e^{\ln(-x)} = e^{-W\left(\ln\sqrt2\right)}\\ \Leftrightarrow -x = e^{-W\left(\ln\sqrt2\right)}\\ \Leftrightarrow x = -e^{-W\left(\ln\sqrt2\right)}\\ \Leftrightarrow x = -e^{-W_o\left(\ln\sqrt2\right)}\\ \approx -0.766664695962123093111204422510314848006675346669832058460884376\dots \end{array}$

 

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