Bài 7:
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(A=\left(\dfrac{3x}{x-2}-\dfrac{2x^2-5}{x^2-4}-\dfrac{x-1}{x+2}\right):\dfrac{3}{x+2}\)
\(=\left(\dfrac{3x}{x-2}-\dfrac{2x^2-5}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-1}{x+2}\right)\cdot\dfrac{x+2}{3}\)
\(=\dfrac{3x\left(x+2\right)-2x^2+5-\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{3}\)
\(=\dfrac{3x^2+6x-2x^2+5-\left(x^2-3x+2\right)}{x-2}\cdot\dfrac{1}{3}\)
\(=\dfrac{x^2+6x+5-x^2+3x-2}{x-2}\cdot\dfrac{1}{3}=\dfrac{9x+3}{3\left(x-2\right)}=\dfrac{3x+1}{x-2}\)
b: \(x^2-2x=0\)
=>x(x-2)=0
=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào A, ta được:
\(A=\dfrac{3\cdot0+1}{0-2}=\dfrac{1}{-2}=-\dfrac{1}{2}\)
c: Để A nguyên thì \(3x+1⋮x-2\)
=>\(3x-6+7⋮x-2\)
=>\(7⋮x-2\)
=>\(x-2\in\left\{1;-1;7;-7\right\}\)
=>\(x\in\left\{3;1;9;-5\right\}\)
Bài 8:
a: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{x+4\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{2x-2+4x+8}{\left(x-2\right)}\cdot\dfrac{x}{3\left(x+1\right)}\)
\(=\dfrac{6x+6}{3\left(x+1\right)}\cdot\dfrac{x}{x-2}=\dfrac{2x}{x-2}\)
b: |x|=1
=>\(\left[{}\begin{matrix}x=-1\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{2\cdot1}{1-2}=\dfrac{2}{-1}=-2\)
Bài 4:
a.
\(x=2\Rightarrow A=\dfrac{2^2-9}{3\left(2+5\right)}=\dfrac{-5}{21}\)
b.
\(B=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-\left(3x^2+9\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
c.
\(P=AB=\dfrac{x^2-9}{3\left(x+5\right)}.\dfrac{3}{x+3}=\dfrac{3\left(x-3\right)\left(x+3\right)}{3\left(x+5\right)\left(x+3\right)}=\dfrac{x-3}{x+5}\)
\(P=\dfrac{x+5-8}{x+5}=1-\dfrac{8}{x+5}\)
\(P\in Z\Rightarrow\dfrac{8}{x+5}\in Z\)
\(\Rightarrow x+5=Ư\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Rightarrow x=\left\{-13;-9;-7;-6;-4;-3;-1;3\right\}\)
Kết hợp ĐKXĐ \(\Rightarrow x=\left\{-13;-9;-7;-6;-4;-1\right\}\)
Bài 3:
a.
ĐKXĐ: \(x\ne\left\{3;-2;2\right\}\)
\(Q=\dfrac{x-1}{x+2}+\dfrac{4x+4}{x^2-4}-\dfrac{3}{x-2}\)
\(=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{4x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x^2-3x+2\right)+4x+4-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b.
\(P=3\Rightarrow\dfrac{x+2}{x-3}=3\Rightarrow x+2=3x-9\)
\(\Rightarrow2x=11\Rightarrow x=\dfrac{11}{2}\)
\(\Rightarrow Q=\dfrac{\dfrac{11}{2}}{\dfrac{11}{2}+2}=\dfrac{11}{15}\)
c.
\(M=PQ=\dfrac{x+2}{x-3}.\dfrac{x}{x+2}=\dfrac{x}{x-3}=\dfrac{x-3+3}{x-3}=1+\dfrac{3}{x-3}\)
M nguyên \(\Rightarrow\dfrac{3}{x-3}\) nguyên
\(\Rightarrow x-3=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x=\left\{0;2;4;6\right\}\)
Kết hợp ĐKXĐ \(\Rightarrow x=\left\{0;4;6\right\}\)
Bài 5:
a.
\(A=\dfrac{2x}{x+3}+\dfrac{2}{x-3}-\dfrac{x^2-x+6}{x^2-9}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+2\left(x+3\right)-\left(x^2-x+6\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x+3}\)
b.
\(A=\dfrac{x}{x+3}=\dfrac{x+3-3}{x+3}=1-\dfrac{3}{x+3}\)
\(A\in Z\Rightarrow\dfrac{3}{x+3}\in Z\)
\(\Rightarrow x+3=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x=\left\{-6;-4;-2;0\right\}\)
Bài 6:
a.
\(A=\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}-\dfrac{x^2+4x}{x^2-4}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-\left(x^2+4x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+3x+2+x^2-3x+2-x^2-4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-2}{x+2}\)
b.
\(x=4\Rightarrow A=\dfrac{4-2}{4+2}=\dfrac{1}{3}\)
c.
\(A=\dfrac{x-2}{x+2}=\dfrac{x+2-4}{x+2}=1-\dfrac{4}{x+2}\)
\(A\in Z\Rightarrow\dfrac{4}{x+2}\in Z\)
\(\Rightarrow x+2=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x=\left\{-6;-4;-3;-1;0;2\right\}\)
Kết hợp ĐKXĐ \(\Rightarrow x=\left\{-6;-4;-3;-1;0\right\}\)
Thay vào A ta thấy chỉ có \(x=\left\{-6;-4;-3\right\}\) cho A nguyên dương
Bài 8:
a.
\(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\left(\dfrac{6x+6}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6x\left(x+1\right)\left(x+2\right)}{3\left(x+1\right)\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x-2}\)
b.
\(\left|x\right|=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Với \(x=-1\Rightarrow A=\dfrac{2.\left(-1\right)}{-1-2}=\dfrac{2}{3}\)
Với \(x=1\Rightarrow A=\dfrac{2.1}{1-2}=-2\)