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Câu 1:
\(Tacó\)
\(\frac{2}{2x-1}+\frac{4x^2+1}{4x^2-1}-\frac{1}{2x+1}=\frac{2}{2x-1}+\frac{4x^2+1}{\left(2x+1\right)\left(2x-1\right)}-\frac{1}{2x+1}\)
\(=\frac{4x+2}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x^2+1}{\left(2x+1\right)\left(2x-1\right)}-\frac{2x-1}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\frac{4x+2+4x^2+1-2x+1}{\left(2x+1\right)\left(2x-1\right)}=\frac{2x\left(2x+1\right)+4}{\left(2x+1\right)\left(2x-1\right)}=\frac{2x+4}{2x-1}\)
\(b,x=\frac{1}{2}\Rightarrow2x-1=0\left(loại\right)\)
..... 2 câu sau easy
\(\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)
\(\Leftrightarrow\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\right):\frac{2\left(x-1\right)}{x}\)
\(\frac{\left(x-3\right)\left(x-3\right)-x^2+9}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\Leftrightarrow\frac{x^2-6x+9-x^2+9}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\frac{18-6x}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\Leftrightarrow\frac{2\left(3-x\right)}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\Leftrightarrow-\left(x-1\right)\)
b) Để A=2 hay \(-\left(x-1\right)=2\)
\(\Rightarrow-x+1=2\Rightarrow x=-1\)
Vậy....
a) \(A=\frac{2x}{x^2-9}+\frac{5}{3-x}-\frac{1}{x+3}\)
\(\Leftrightarrow A=\frac{2x}{\left(x-3\right)\left(x+3\right)}-\frac{5}{x-3}-\frac{1}{x+3}\)
\(\Leftrightarrow A=\frac{2x-5\left(x+3\right)-x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x-5x-15-x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-4x-12}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4}{x-3}\)
b) x = \(\frac{-3}{2}\) ( thỏa mãn )
Vậy với x= \(\frac{-3}{2}\) thì giá trị của biểu thức A bằng \(\frac{-4}{\frac{-3}{2}-3}=\frac{-4}{\frac{-9}{2}}=\left(-4\right)\left(\frac{-2}{9}\right)=\frac{8}{9}\)
c) A là số nguyên
\(\Leftrightarrow\frac{-4}{x-3}\)nguyên
\(\Leftrightarrow x-3=Ư\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng sau :
Vậy.................