Giải phương trình ` x^2 = 2^x` Chỉ nhận bài làm có cách giải cụ thể nhé !!!!!!

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