Question

Bài 1

A=\(\frac{x+2}{x}\) ;B=\(\frac{x^2}{x^2-4}+\frac{1}{x-2}+\frac{1}{x+2}\) ĐKXĐ: x \(\ne\) 2 ; x > 0

a)Rút gọn B và tính P=\(\frac{A}{B}\)

b)Tính x để B =giá trị tuyệt đối của 13

Answer 1

\(ĐKxĐ:x\ne\pm2;x\ne0\)

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

\(P=\frac{A}{B}=\frac{x+2}{x}:\frac{x}{x-2}=\frac{\left(x+2\right)\left(x-2\right)}{x^2}=\frac{x^2-4}{x^2}\)

\(b,B=\frac{x}{x-2}=\left|13\right|=13\Leftrightarrow x=13\left(x-2\right)\Leftrightarrow x=13x-26\Leftrightarrow12x-26=0\Leftrightarrow x=\frac{26}{12}=\frac{13}{6}\)