a/ \(\left|5x-|x-2|\right|=0\)
\(\Leftrightarrow5x-\left|x-2\right|=0\)
\(\Leftrightarrow\left|x-2\right|=5x\)
+) TH1: Với x - 2 ≥ 0 <=> x ≥ 2 có:
x - 2 = 5x <=> -4x = 2 <=> \(x=-\dfrac{1}{2}\) (ktm)
+) TH2: Với x - 2 < 0 <=> x < 2 có:
x - 2 = -5x <=> 6x = 2 <=> x = \(\dfrac{1}{3}\left(tm\right)\)
Vậy pt có 1 nghiệm x = 1/3
b/ \(x^3-27=x^2+3x+9\)
\(\Leftrightarrow x^3-x^2-3x-36=0\)
\(\Leftrightarrow x^3+3x^2-4x^2+9x-12x-36=0\)
\(\left(x^3+3x^2+9x\right)+\left(-4x^2-12x-36\right)=0\)
\(\Leftrightarrow x\left(x^2+3x+9\right)-4\left(x^2+3x+9\right)=0\)
\(\Leftrightarrow\left(x^2+3x+9\right)\left(x-4\right)=0\)
Vì: \(x^2+3x+9=\left(x^2+2\cdot1,5\cdot x+2,25\right)+6,75=\left(x+1,5\right)^2+6,75\ge6,75>0\) (vô nghiệm)
nên: x- 4 = 0 <=> x = 4
Vậy pt có 1 nghiệm x= 4
