Question

P=\(\sqrt{9x^2-6x+1}+1-4x\)

tìm x>1 sao cho P= -4

Answer 1

\(P=-4\Leftrightarrow\sqrt{\left(3x-1\right)^2}+1-4x=-4\\ \Leftrightarrow\left|3x-1\right|=4x-5\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=4x-5\left(x\ge\dfrac{1}{3}\right)\\3x-1=5-4x\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{6}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=4\)

Answer 2

P=\(\sqrt{\left(3x-1\right)^2}+1-4x\)

P= 3x-1+1-4x 

P= -x    mà P= -4

=> x=4