Bài làm

\(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\)( có sửa )

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm2\\x\ne4\end{cases}}\)

\(=\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3x-6}+\frac{1}{x+2}\)

\(=\frac{x}{x^2-4}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\)

\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\)

\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2x+4}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\)

Nếu mình sửa đề sai thì ib mình làm lại cho

MK nghĩ là C

theo mình là

C nhé

\(M=\frac{x+2xy-2y-1}{4y^2-1}\div\frac{3x^2}{2y-1}\)

\(M=\frac{x\left(2y+1\right)-\left(2y+1\right)}{\left(2y-1\right)\left(2y+1\right)}.\frac{2y-1}{3x^2}\)

\(M=\frac{\left(x-1\right)\left(2y+1\right)\left(2y-1\right)}{3x^2\left(2y-1\right)\left(2y+1\right)}=\frac{x-1}{3x^2}\)không phụ thuộc vào biến \(y\)

\(M=\frac{x+2xy-2y-1}{4y^2-1}:\frac{3x^2}{2y-1}=\frac{x\left(2y+1\right)-\left(2y+1\right)}{\left(2y-1\right)\left(2y+1\right)}:\frac{3x^2}{2y-1}\)

\(=\frac{\left(x-1\right)\left(2y+1\right)}{\left(2y-1\right)\left(2y+1\right)}:\frac{3x^2}{2y-1}=\frac{x-1}{2y-1}.\frac{2y-1}{3x^2}=\frac{x-1}{3x^2}\)

Vậy giá trị biểu thức ko phụ thuộc biến y