\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \Leftrightarrow6-5x-4x^2=x^2+4x+4\\ \Leftrightarrow5x^2+9x-2=0\\ \Leftrightarrow5\left(x+\dfrac{9}{10}\right)^2=\dfrac{121}{20}\\ \Leftrightarrow\left(x+\dfrac{9}{10}\right)^2=\dfrac{\dfrac{121}{20}}{5}=\dfrac{121}{100}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{9}{10}=\dfrac{11}{10}\\x+\dfrac{9}{10}=-\dfrac{11}{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2\end{matrix}\right.\)
vậy x cần tìm là 0,2 và 2
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