\(a,\sqrt{4x^2+4x+1}-5x=2\)
=> \(\sqrt{\left(2x+1\right)^2}-5x=2\)
=>\(\left|2x+1\right|=2+5x\)
=> \(\left[{}\begin{matrix}2x+1=2+5x\\2x+1=-\left(2+5x\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x-5x=2-1\\2x+1=-2x-5x\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}-3x=1\\2x+5x=-2-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{1}{3}\\7x=-3=>x=\frac{3}{7}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};\frac{3}{7}\right\}\)
\(b,\sqrt{x^2-x+\frac{1}{4}}=5-x\)
=> \(\sqrt{\left(x-\frac{1}{2}\right)^2}=5-x\)
=> \(\left|x-\frac{1}{2}\right|=5-x\)
=> \(\left[{}\begin{matrix}x-\frac{1}{2}=5-x\\x-\frac{1}{2}=-\left(5-x\right)=-5+x\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+x=5+\frac{1}{2}\\x+\frac{-1}{2}=-5+x\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}2x=\frac{11}{2}=>x=\frac{11}{4}\\-\frac{1}{2}=-5\left(sai\right)=>x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{11}{4}\right\}\)
Bình phương sẽ lẹ hơn sử dụng hằng đẳng thức rồi phá trị tuyệt đối chia trường hợp
a/ \(\Leftrightarrow\sqrt{4x^2+4x+1}=5x+2\) (\(x\ge-\frac{2}{5}\))
\(\Leftrightarrow4x^2+4x+1=25x^2+20x+4\)
\(\Leftrightarrow21x^2+16x+3=0\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-\frac{3}{7}\left(l\right)\end{matrix}\right.\)
b/ ĐK: \(x\le5\)
\(x^2-x+\frac{1}{4}=x^2-10x+25\)
\(\Leftrightarrow9x=\frac{99}{4}\Rightarrow x=\frac{11}{4}\)
Em thử nhé, sai thì thôi ạ :(
a) ĐK: \(x\in\mathbb{R}\)
\(PT\Leftrightarrow\left|2x+1\right|-5x=2\)
Với \(x< -\frac{1}{2}\Rightarrow2x+1< 0\). Khi đó PT trở thành:
\(-2x-1-5x=2\Leftrightarrow x=-\frac{3}{7}\left(KTM\right)\)
Với x>= -1/2 thì tương tự
