Question

[Diệu Khói]

 *Tìm điều kiện xác định và rút gọn biểu thức

a) Q= (\(\frac{1}{X-1}\) - \(\frac{1}{X}\)) : (\(\frac{X+1}{X-2}\)- \(\frac{X+2}{X-1}\))

b) C= (\(\frac{x+2}{x^2-x}\) + \(\frac{x-2}{x^2+x}\)) .\(\frac{x^2-1}{x^2+2}\)

Answer 1

[Quỳnh'ss Ok'ss]

Bài làm

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

\(Q=\left(\frac{x}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}\right):\left(\frac{x^2-1}{\left(x-2\right)\left(x-1\right)}-\frac{x^2-4}{\left(x-2\right)\left(x-1\right)}\right)\)

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

\(Q=\frac{1}{x\left(x-1\right)}:\frac{3}{\left(x-2\right)\left(x-1\right)}\)

\(Q=\frac{1}{x\left(x-1\right)}.\frac{\left(x-2\right)\left(x-1\right)}{3}\)

\(Q=\frac{x-2}{3x}\)

ĐKXĐ: \(\frac{x-2}{3}\ge0\)

Vì \(\frac{x-2}{3}\ge0\). Mà 3 > 0

=> x - 2 > 0

<=> x > 2

Vậy x > 2 thì biểu thức Q có nghĩa.

b) \(C=\left(\frac{x+2}{x^2-x}+\frac{x-2}{x^2+x}\right).\frac{x^2-1}{x^2+2}\)

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

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

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

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

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

\(C=\frac{2}{x}\)
ĐKXĐ: \(\frac{2}{x}\ge0\)

Vì \(\frac{2}{x}\ge0\),

Mà 2 > 0

=> x > 0 

Vậy x > 0 thì biểu thức C có nghĩa.