Bài 6: \(\left(\frac{2x}{3x+1}-1\right):\left(1-\frac{8x^2}{9x^2-1}\right)\)
\(=\frac{2x-3x-1}{3x+1}:\frac{9x^2-1-8x^2}{9x^2-1}\)
\(=\frac{-\left(x+1\right)}{3x+1}\cdot\frac{\left(3x-1\right)\left(3x+1\right)}{x^2-1}=\frac{-\left(x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-3x+1}{x-1}\)
BÀi 7:
\(D=\frac{3x-2}{9x^2-4}\)
\(=\frac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\frac{1}{3x+2}\)
Thay x=1/3 vào D, ta được:
\(D=\frac{1}{3\cdot\frac13+2}=\frac{1}{1+2}=\frac13\)
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