Bài 1 :
a/ ĐKXĐ : \(a\ne0;-1\)
Ta có :
\(M=\left(\frac{1}{a}+\frac{a}{a+1}\right)-\frac{a}{a^2+a}\)
\(=\left(\frac{a+1}{a\left(a+1\right)}-\frac{a^2}{a\left(a+1\right)}\right)-\frac{a}{a\left(a+1\right)}\)
\(=\frac{a-a^2+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}\)
\(=\frac{1-a^2}{a\left(a+1\right)}\)
\(=\frac{\left(1-a\right)\left(1+a\right)}{a\left(a+1\right)}\)
\(=\frac{1-a}{a}\)
Vậy....
c/ Ta có : \(a+1=0\Leftrightarrow a=-1\) (loại)
Vậy....
Bài 2 :
a/ ĐKXĐ : \(x\ne0;3;-3\)
Ta có :
\(A=\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x-3}\)
\(=\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x-3}{x}\)
\(=\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{x}\)
\(=\frac{x^2+x}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{x}\)
\(=\frac{x\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{x}\)
\(=\frac{x+1}{x+3}\)
Vậy....
b/ \(A=3\)
\(\Leftrightarrow\frac{x+1}{x+3}=3\)
\(\Leftrightarrow x+1=3x+9\)
\(\Leftrightarrow2x=-8\Leftrightarrow x=-4\)
Vậy...
a) ĐKXĐ: \(a\notin\left\{0;-1\right\}\)
Ta có: \(M=\left(\frac{1}{a}+\frac{a}{a+1}\right)-\frac{a}{a^2+a}\)
\(=\frac{a+1}{a\left(a+1\right)}+\frac{a^2}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}\)
\(=\frac{a+1+a^2-a}{a\left(a+1\right)}\)
\(=\frac{a^2+1}{a\left(a+1\right)}\)
b) Ta có: a+1=0
hay a=-1(loại)
Vậy: Khi a+1=0 thì M không có giá trị
Bài 2:
a) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(A=\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x-3}\)
\(=\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x-3}{x}\)
\(=\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x}\)
\(=\frac{x^2+x}{x+3}\cdot\frac{1}{x}\)
\(=\frac{x\left(x+1\right)}{\left(x+3\right)\cdot x}\)
\(=\frac{x+1}{x+3}\)
b) Để A=3 thì \(\frac{x+1}{x+3}=3\)
\(\Leftrightarrow x+1=3\left(x+3\right)\)
\(\Leftrightarrow x+1=3x+9\)
\(\Leftrightarrow x+1-3x-9=0\)
\(\Leftrightarrow-2x-8=0\)
\(\Leftrightarrow-2x=8\)
hay x=-4(nhận)
Vậy: Khi A=3 thì x=-4
