Đáp án:
Bài 1:
\(a)\dfrac{{4ab}}{{3c}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{3.4.{a^2}{b^3}c}}{{3.3.a{b^2}{c^2}}} = \dfrac{{4ab}}{{3c}}\\
b)\dfrac{{4.2.2.{x^5}{y^6}z}}{{4.2.{x^2}y{z^4}}} = \dfrac{{2{x^3}{y^5}}}{{{z^3}}}\\
c)\dfrac{{3x{{\left( {x - y} \right)}^3}}}{{2{x^2}{{\left( {x - y} \right)}^2}}} = \dfrac{{3\left( {x - y} \right)}}{{2x}}\\
d)\dfrac{{15a\left( {a - 1} \right)}}{{10ab\left( {1 - a} \right)}} = - \dfrac{{3\left( {a - 1} \right)}}{{2b\left( {a - 1} \right)}} = - \dfrac{3}{{2b}}\\
B2:\\
a)\dfrac{{{{\left( {2x + 3} \right)}^2}}}{{\left( {x - 2} \right)\left( {2x + 3} \right)}} = \dfrac{{2x + 3}}{{x - 2}}\\
b)\dfrac{{ - \left( {{x^2} - 2xy + {y^2}} \right) + {z^2}}}{{{x^2} + 2xz + {z^2} + {y^2} - 2{z^2}}}\\
= \dfrac{{ - {{\left( {x - y} \right)}^2} + {z^2}}}{{{{\left( {x + z} \right)}^2} + {y^2} - {z^2} - {z^2}}}\\
= \dfrac{{\left( {z - z + y} \right)\left( {z + x - y} \right)}}{{\left( {x + z - z} \right)\left( {x + z + z} \right) + \left( {y - z} \right)\left( {y + z} \right)}}\\
= \dfrac{{\left( {z - z + y} \right)\left( {z + x - y} \right)}}{{x\left( {x + 2z} \right) + \left( {y - z} \right)\left( {y + z} \right)}}
\end{array}\)
