Question

[Ly Khánh Trần]

bài 2: tìm x:
a) (4x4+3x3)÷(−x3)+(15x2+6x)÷3x=0
b)(x2−12x)÷2x−(3x−1)2÷(3x−1)=0

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 2: Tìm x

a)ĐKXĐ: \(x\ne0\)

Ta có: \(\left(4x^4+3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow\frac{-x^3\left(4x+3\right)}{x^3}+\frac{3x\left(5x+2\right)}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: x=1

b) ĐKXĐ: \(x\notin\left\{0;\frac{1}{3}\right\}\)

Ta có: \(\left(x^2-12x\right):2x-\left(3x-1\right)^2:\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{x\left(x-12\right)}{2x}-\frac{\left(3x-1\right)^2}{\left(3x-1\right)}=0\)

\(\Leftrightarrow\frac{x-12}{x}-3x+1=0\)

\(\Leftrightarrow\frac{x-12}{x}=3x-1\)

\(\Leftrightarrow x-12=x\left(3x-1\right)\)

\(\Leftrightarrow3x^2-x+x-12=0\)

\(\Leftrightarrow3x^2-12=0\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-2\right\}\)