Question

1: tìm x: \(\left|x-2\right|-2\left|3x+6\right|=5-2\left(1-x\right)\)

2: giải: a) \(8x^2-2x-1=0\)

Answer 1

2)

\(8x^2-2x-1=0\)

\(\Leftrightarrow 8x^2-4x+2x-1=0\)

\(\Leftrightarrow 4x(2x-1)+(2x-1)=0\)

\(\Leftrightarrow (2x-1)(4x+1)=0\)

\(\Leftrightarrow \left[\begin{matrix} 2x-1=0\rightarrow x=\frac{1}{2}\\ 4x+1=0\rightarrow x=\frac{-1}{4}\end{matrix}\right.\)

Answer 2

1)

Nếu \(x\geq 2\Rightarrow |x-2|=x-2; |3x+6|=3x+6\)

Khi đó, pt trở thành:

\(x-2-2(3x+6)=5-2(1-x)\)

\(\Leftrightarrow -5x-14=3+2x\Leftrightarrow x=\frac{-17}{7}\) (loại vì \(x\geq 2\) )

Nếu \(2> x\geq -2\Rightarrow |x-2|=2-x; |3x+6|=3x+6\)

Khi đó pt trở thành:

\(2-x-2(3x+6)=5-2(1-x)\)

\(\Leftrightarrow -10-7x=2x+3\)

\(\Leftrightarrow x=-\frac{13}{9}\) (thỏa mãn)

Nếu \(x< -2\Rightarrow |x-2|=2-x; |3x+6|=-(3x+6)\)

Khi đó pt trở thành:
\(2-x+2(3x+6)=5-2(1-x)\)

\(\Leftrightarrow 5x+14=3+2x\Leftrightarrow x=\frac{-11}{3}\) (thỏa mãn)

Vậy \(x=\frac{-11}{3}; x=\frac{-13}{9}\)