Question

1. Giải pt

b. x\(^3\)-16x\(^2\)+64x= 0

c.|6x-1|= 2x+5

d. |5-x|= 8x-10

e. \(\dfrac{x-1}{x+3}\)= \(\dfrac{x}{x-3}\)+ \(\dfrac{12}{x^2-9}\)

g. \(\dfrac{x+2}{x-2}\)- \(\dfrac{x-2}{x+2}\)= \(\dfrac{6}{x^2-4}\)

Answer 1

b.\(x^3-16x^2+64x=0\)

=>\(x^3-8x^2-8x^2+64x=0\)

=>\(x^2\left(x-8\right)-8x\left(x-8\right)=0\)

=>\(x\left(x-8\right)\left(x-8\right)=0\)

=>\(x=0\) và \(x-8=0\)

=>x=0 và x= 8

Vậy S={0; 8}

Answer 2

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

-Nếu 6x - 1 \(\ge0\Leftrightarrow x\ge\dfrac{1}{6}\)

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

\(\Leftrightarrow6x-1=2x+5\)

\(\Leftrightarrow6x-2x=5+1\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\dfrac{3}{2}\) (Loại)

-Nếu 6x-1 < 0 \(\Leftrightarrow x< \dfrac{1}{6}\)

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

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

\(\Leftrightarrow-6x-2x=5-1\)

\(\Leftrightarrow-8x=4\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)(Nhận)

Vậy S={\(-\dfrac{1}{2}\)}

Answer 3

e.\(\dfrac{x-1}{x+3}=\dfrac{x}{x-3}+\dfrac{12}{x^2-9}\).ĐKXĐ: x\(\ne\pm3\)

\(\Rightarrow x^2-4x+3=x^2+3x+12\)

\(\Leftrightarrow x^2-x^2-4x-3x=12-3\)

\(\Leftrightarrow-7x=9\)

\(\Leftrightarrow x=\dfrac{-9}{7}\)(Thỏa mãn )

Vậy S={\(\dfrac{-9}{7}\)}

Answer 4