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Ta có : \(\frac{x}{x-3}-\frac{x^2+3x}{2x+3}\left(\frac{x+3}{x^2-3x}-\frac{x}{x^2-9}\right)\)
\(=\frac{x}{x-3}-\frac{x^2+3x}{2x+3}\left(\frac{x+3}{x\left(x-3\right)}-\frac{x}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x}{x-3}-\frac{x^2+3x}{2x+3}\left(\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}-\frac{x^2}{\left(x-3\right)\left(x+3\right)x}\right)\)
\(=\frac{x}{x-3}-\frac{x^2+3x}{2x+3}\left(\frac{\left(x+3\right)^2-x^2}{x\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x}{x-3}-\frac{x^2+3x}{2x+3}\left(\frac{x^2+6x+9-x^2}{x\left(x^2-3\right)}\right)\)
\(=\frac{x}{x-3}-\frac{x^2+3x}{2x+3}\left(\frac{3\left(2x+3\right)}{x\left(x^2-3\right)}\right)\)
\(=\frac{x}{x-3}-\frac{3x^2+9x}{x\left(x^2-3\right)}\)(mk sợ mk làm sai lắm nếu làm sai thì sory nhá)
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
sắp thi hkì r chỉ jùm vs ik
A=(\(\frac{x^3-1}{x\left(x-1\right)}\)-\(\frac{x^3-1}{x\left(x+1\right)}\)) : \(\frac{2\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\)ĐKXĐ: x\(\ne\) -1, 1
A=\(\frac{1}{x\left(x+1\right)}\)x \(\frac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-1\right)}\)
A=\(\frac{1}{2x^2-2x}\)
B=\(\frac{x+1}{x-2}\)-\(\frac{2x}{x+2}\)-\(\frac{2+5x}{x^2-4}\)ĐKXĐ : x\(\ne\)2, -2
B=\(\frac{x+1}{x-2_{ }}\)-\(\frac{2x}{x+2}\)-\(\frac{2+5x}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}\)-\(\frac{2x^2-4x}{\left(x-2\right)\left(x+2\right)}\)-\(\frac{2+5x}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{-x^2+2x}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{-x}{x+2}\)