Question

[ZzHxHzZ]

Bài 1: Thực hiện phép tính
a, \(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\)

Bài 2: Với giá trị nào của x thì giá trị của mỗi biểu thức sau có giá trị bằng 0

a, \(\dfrac{5}{x-2}-\dfrac{1}{x+2}+\dfrac{4}{x^2}\)

b, \(\dfrac{2}{x^2-x+1}+x+1\)

Answer 1

[Mysterious Person]

bài a) \(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\)

= \(\dfrac{x.x+\left(x-1\right)\left(x+1\right)}{\left(x+1\right)x}:\dfrac{x.x-\left(x-1\right)\left(x+1\right)}{\left(x+1\right)x}\)

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

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

Answer 2

[Mysterious Person]

bài 2) ta có mỗi biểu thức sau bằng 0

a) \(\Leftrightarrow\) \(\dfrac{5}{x-2}-\dfrac{1}{x+2}+\dfrac{4}{x^2}=0\)

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

\(\Leftrightarrow\)\(\dfrac{5x+10-x+2}{x^2-4}+\dfrac{4}{x^2}=0\)

\(\Leftrightarrow\) \(\dfrac{5x^2-x+12}{x^2-4}+\dfrac{4}{x^2}=0\)

\(\Leftrightarrow\) \(\dfrac{\left(5x^2-x+12\right)x^2+4\left(x^2-4\right)}{\left(x^2-4\right)x^2}=0\)

\(\Leftrightarrow\) \(\dfrac{5x^4-x^3+12x^2+4x^2-16}{x^4-4x^2}=0\)