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\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}-\frac{2\left(-x-1\right)}{x+1}\right]:\frac{x-1}{x}\)
\(=\)\(\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}+\frac{2\left(x+1\right)}{x+1}\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)
\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\)\(\left(đpcm\right)\)
b) \(\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-\left(x+1\right)\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(x+1\right)\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-2\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)
\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\left(đpcm\right)\)
a) \(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left(\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\left(\frac{9}{x\left(x+3\right)\left(x-3\right)}+\frac{x^2-3x}{x\left(x+3\right)\left(x-3\right)}\right)\)
\(:\left(\frac{3x-9}{3x\left(x+3\right)}-\frac{x^2}{3x\left(x+3\right)}\right)\)
\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}:\frac{-x^2+3x-9}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\frac{x^2-3x+9}{x-3}.\frac{3}{x^2+3x-9}\)
\(=\frac{x^2-3x+9}{3-x}.\frac{3}{x^2-3x+9}\)
\(=\frac{3}{3-x}\left(đpcm\right)\)
tổng 2 số là 150, tổng của 1/6 số này và 1/9 số kia = 18. Tìm 2 số đó
điều kiện \(x\ne2;x\ne-2\)
\(\frac{\left(x+2\right)^2.P}{x^2-4}=\frac{\left(x-1\right).Q}{x^2-4}\)
\(\left(x+2\right)^2.P=\left(x-1\right)Q\)
\(P=x-1\)
\(Q=\left(x+2\right)^2\)
\(=x^3-\left(\frac{1}{2}\right)^3=x^3-\frac{1}{8}\)
\(\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)\)
\(=x.\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)-\frac{1}{2}.\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)\)
\(=x^3+\frac{1}{2}x^2+\frac{1}{4}x-\frac{1}{2}x^2-\frac{1}{4}x-\frac{1}{8}\)
\(=x^3-\frac{1}{8}\)
Làm bằng cách hằng đẳng thức còn rắc rối hơn thà nhân lên luôn.