Đề bài

Thực hiện phép tính: \(a)\left( { - \frac{{3{\rm{x}}}}{{5{\rm{x}}{y^2}}}} \right):\left( { - \frac{{5{y^2}}}{{12{\rm{x}}y}}} \right)\)

\(b)\frac{{4{{\rm{x}}^2} - 1}}{{8{{\rm{x}}^3} - 1}}:\frac{{4{{\rm{x}}^2} + 4{\rm{x}} + 1}}{{4{{\rm{x}}^2} + 2{\rm{x}} + 1}}\)

Lời giải chi tiết

\(a)\left( { - \frac{{3x}}{{5x{y^2}}}} \right):\left( { - \frac{{5{y^2}}}{{12xy}}} \right) = \frac{{ - 3x}}{{5x{y^2}}}.\frac{{ - 12xy}}{{5{y^2}}} = \frac{36{x^2y}}{25xy^4} = \frac{36{x}}{25y^3}\) b) \(\frac{4{{x}^{2}}-1}{8{{x}^{3}}-1}:\frac{4{{x}^{2}}+4x+1}{4{{x}^{2}}+2x+1}=\frac{4{{x}^{2}}-1}{8{{x}^{3}}-1}.\frac{4{{x}^{2}}+2x+1}{4{{x}^{2}}+4x+1}\) \(=\frac{\left( 2x-1 \right)\left( 2x+1 \right)\left( 4{{x}^{2}}+2x+1 \right)}{\left( 2x-1 \right)\left( 4{{x}^{2}}+2x+1 \right){{\left( 2x+1 \right)}^{2}}}=\frac{1}{2x+1}\).