Question

[bùi huyền trang]

cho:

P = \(\frac{x+2}{x-1}\); Q =\(\frac{x-1}{x}+\frac{2x+1}{x^2+x}\)

a) tính P khi \(\left|x-2\right|=1\)

b) rút gọn Q

c) tìm số tự nhiên x để M = P : Q
 

Answer 1

[Nguyễn Phương Uyên]

\(ĐKXĐ:x\ne0;x\ne\pm1\)

a. |x - 2| = 1

=> x - 2 = 1 hoặc x - 2 = -1

=> x = 3 hoặc x = 1 (loại)

=> x = 3

\(P=\frac{3+2}{3-1}=\frac{5}{2}\)

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

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

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

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

M đâu ra

Answer 2

[Minh Nguyen]

a) \(ĐKXĐ:x=1\)

Để \(\left|x-2\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=1\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow x=3\)

\(\Leftrightarrow P=\frac{3+2}{3-1}=\frac{5}{2}\)

b) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

\(Q=\frac{x-1}{x}+\frac{2x+1}{x^2+x}\)

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

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

\(\Leftrightarrow Q=\frac{x^2+2x}{x\left(x+1\right)}\)

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

\(\Leftrightarrow Q=\frac{x+2}{x+1}\)

c) \(ĐKXĐ:x\ne\pm1\)

 \(M=P:Q\)

\(\Leftrightarrow M=\frac{x+2}{x-1}:\frac{x+2}{x+1}\)

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

\(\Leftrightarrow M=\frac{x+1}{x-1}\)

Đoạn này mik k hiểu tìm x để làm j ?