Bài 11:
a) Thay \(x=4\) vào A ta có:
\(A=\dfrac{4+2}{4+1}=\dfrac{6}{5}\)
b) \(B=\dfrac{3}{x-1}-\dfrac{x+5}{x^2-1}\)
\(B=\dfrac{3}{x-1}-\dfrac{x+5}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+5}{\left(x-1\right)\left(x+1\right)}\)
\(B=\dfrac{3x+3-x-5}{\left(x+1\right)\left(x-1\right)}\)
\(B=\dfrac{2x-2}{\left(x+1\right)\left(x-1\right)}\)
\(B=\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\dfrac{2}{x+1}\left(dpcm\right)\)
c) Ta có: \(A-B=0\) khi:
\(\dfrac{x+2}{x+1}-\dfrac{2}{x+1}=0\) (ĐK: \(x\ne-1\))
\(\Leftrightarrow\dfrac{x+2-2}{x+1}=0\)
\(\Leftrightarrow\dfrac{x}{x+1}=0\)
\(\Leftrightarrow x=0\left(tm\right)\)
Bài 10:
a) Thay \(x=-2\) vào A ta có:
\(A=\dfrac{-2+2}{-2-5}=\dfrac{0}{-7}=0\)
b) \(B=\dfrac{3}{x+5}+\dfrac{20-2x}{x^2-25}\)
\(B=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{20-2x}{\left(x+5\right)\left(x-5\right)}\)
\(B=\dfrac{3x-15+20-2x}{\left(x+5\right)\left(x-5\right)}\)
\(B=\dfrac{x+5}{\left(x+5\right)\left(x-5\right)}\)
\(B=\dfrac{1}{x-5}\left(dpcm\right)\)
c) Ta có:
\(A=B\) khi:
\(\dfrac{x+2}{x-5}=\dfrac{1}{x-5}\) (ĐK: \(x\ne5\))
\(\Leftrightarrow\dfrac{x+2}{x-5}-\dfrac{1}{x-5}=0\)
\(\Leftrightarrow\dfrac{x+2-1}{x-5}=0\)
\(\Leftrightarrow\dfrac{x+1}{x-5}=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
11:
a: Khi x=4 thì A=(4+2)/(4+1)=6/5
b: \(B=\dfrac{3x+3-x-5}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x-2}{\left(x+1\right)\left(x-1\right)}=\dfrac{2}{x+1}\)
c: A-B=0
=>A=B
=>2/(x+1)=(x+2)/(x+1)
=>x+2=2
=>x=0
