Bài tập 1 trang 168 SGK Đại số & Giải tích 11

Áp dụng công thức: \(\left ( \frac{u}{v} \right )'=\frac{u'v-uv'}{v^2},(v(x) \ne 0)\).

Câu a:

\(\begin{array}{l}
y' = \frac{{\left( {x - 1} \right)'.\left( {5x - 2} \right) - \left( {x - 1} \right).\left( {5x - 2} \right)'}}{{{{\left( {5x - 2} \right)}^2}}}\\
 = \frac{{5x - 2 - \left( {x - 1} \right).5}}{{{{\left( {5x - 2} \right)}^2}}} = \frac{3}{{{{\left( {5x - 2} \right)}^2}}}
\end{array}\)

Câu b:

 \(\begin{array}{l}
y' = \frac{{\left( {2x + 3} \right)'.\left( {7 - 3x} \right) - \left( {2x + 3} \right).\left( {7 - 3x} \right)'}}{{{{\left( {7 - 3x} \right)}^2}}}\\
 = \frac{{2\left( {7 - 3x} \right) - \left( {2x + 3} \right).\left( { - 3} \right)}}{{{{\left( {7 - 3x} \right)}^2}}} = \frac{{23}}{{{{\left( {7 - 3x} \right)}^2}}}
\end{array}\)

Câu c:

 \(\begin{array}{l}
y' = \frac{{\left( {{x^2} + 2x + 3} \right)'.\left( {3 - 4x} \right) - \left( {{x^2} + 2x + 3} \right).\left( {3 - 4x} \right)'}}{{{{\left( {3 - 4x} \right)}^2}}}\\
 = \frac{{\left( {2x + 2} \right).\left( {3 - 4x} \right) - \left( {{x^2} + 2x + 3} \right).\left( { - 4} \right)}}{{{{\left( {3 - 4x} \right)}^2}}} = \frac{{ - 2\left( {2{x^2} - 3x - 9} \right)}}{{{{\left( {3 - 4x} \right)}^2}}}
\end{array}\)

Câu d:

\(\begin{array}{l}
y' = \frac{{\left( {{x^2} + 7x + 3} \right)'.\left( {{x^2} - 3x} \right) - \left( {{x^2} + 7x + 3} \right).\left( {{x^2} - 3x} \right)'}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\
 = \frac{{\left( {2x - 7} \right).\left( {{x^2} - 3x} \right) - \left( {{x^2} + 7x + 3} \right).\left( {2x - 3} \right)}}{{{{\left( {{x^2} - 3x} \right)}^2}}} = \frac{{ - 10{x^2} - 6x + 9}}{{{{\left( {{x^2} - 3x} \right)}^2}}}
\end{array}\)

-- Mod Toán 11 HỌC247