Question

Rút gọn biểu thức:\(A=\dfrac{6x+4-x^2}{x^2-4}+\dfrac{x-1}{x+2}+\dfrac{1}{2-x}\)

Answer 1

\(A=\dfrac{6x+4-x^2}{x^2-4}+\dfrac{x-1}{x+2}+\dfrac{1}{2-x}\)

\(\Leftrightarrow A=\dfrac{6x+4-x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-1}{x+2}-\dfrac{1}{x-2}\) MTC: \(\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow A=\dfrac{6x+4-x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{6x+4-x^2+\left(x-1\right)\left(x-2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{6x+4-x^2+x^2-2x-x+2-x-2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{2}{x-2}\)

Answer 2

Khỏi chép đề lại nhazz

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

\(\Leftrightarrow A=\dfrac{6x+4-x^2+\left(x-1\right)\left(x-2\right)-x-2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{6x+4-x^2+x^2-3x+2-x-2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x-2}\)

Answer 3

\(A=\dfrac{6x+4-x^2}{x^2-4}+\dfrac{x-1}{x+2}+\dfrac{1}{2-x}\)

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

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

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

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

\(=\dfrac{2}{x-2}\)