Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
7,\\
\int\limits_0^1 {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right)dx} \\
= \int\limits_0^1 {\frac{{dx}}{{x + 1}}} - \int\limits_0^1 {\frac{{dx}}{{x + 2}}} \\
= \mathop {\left. {\left( {\ln \left| {x + 1} \right| - \ln \left| {x + 2} \right|} \right)} \right|}\nolimits_0^1 \\
= \left( {\ln 2 - \ln 3} \right) - \left( {\ln 1 - \ln 2} \right)\\
= 2\ln 2 - \ln 3\\
\Rightarrow a = 2;\,\,\,\,b = - 1\\
\Rightarrow a + 2b = 0\\
15,\\
P\left( {x; - 1; - 1} \right);\,\,\,Q\left( {3; - 3;1} \right)\\
\Rightarrow \overrightarrow {PQ} = \left( {3 - x;\,\, - 2;\,\,2} \right)\\
PQ = 3\\
\Leftrightarrow \sqrt {{{\left( {3 - x} \right)}^2} + {{\left( { - 2} \right)}^2} + {2^2}} = 3\\
\Leftrightarrow {\left( {3 - x} \right)^2} + 4 + 4 = 9\\
\Leftrightarrow {\left( {3 - x} \right)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
3 - x = 1\\
3 - x = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 4
\end{array} \right.
\end{array}\)
