Đáp án:
B8:
a) \(\left\{ \begin{array}{l}
x = \dfrac{1}{3}\\
y = \dfrac{5}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B7:\\
\left\{ \begin{array}{l}
- 2ax + 2y = - 10\\
2ax + 3{a^2}y = 7a
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {3{a^2} + 2} \right)y = 7a - 10\\
x = \dfrac{{y + 5}}{a}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{7a - 10}}{{3{a^2} + 2}}\\
x = \dfrac{{\dfrac{{7a - 10}}{{3{a^2} + 2}} + 5}}{a} = \dfrac{{7a - 10 + 15{a^2} + 10}}{{a\left( {3{a^2} + 2} \right)}} = \dfrac{{15a - 7}}{{3{a^2} + 2}}
\end{array} \right.\\
a)Thay:a = 2\\
\to \left\{ \begin{array}{l}
x = \dfrac{{15.2 - 7}}{{{{3.2}^2} + 2}} = \dfrac{{23}}{{14}}\\
y = \dfrac{{7.2 - 10}}{{{{3.2}^2} + 2}} = \dfrac{4}{{14}}
\end{array} \right.\\
b)Do:3{a^2} + 2 > 0\forall a
\end{array}\)
⇒ Hệ có nghiệm duy nhất với mọi a
\(\begin{array}{l}
B8:\\
\left\{ \begin{array}{l}
\left( {m - 1} \right)x = 1 - 2\\
x + y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{1}{{m - 2}}\\
y = 2 - \left( { - \dfrac{1}{{m - 2}}} \right) = \dfrac{{2m - 4 + 1}}{{m - 2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{1}{{m - 2}}\\
y = \dfrac{{2m - 3}}{{m - 2}}
\end{array} \right.\\
a)Thay:m = - 1\\
\to \left\{ \begin{array}{l}
x = - \dfrac{1}{{ - 1 - 2}} = \dfrac{1}{3}\\
y = \dfrac{{2\left( { - 1} \right) - 3}}{{ - 1 - 2}} = \dfrac{5}{3}
\end{array} \right.\\
b)DK:m - 2 \ne 0\\
\to m \ne 2\\
B9:\\
DK:x > 0;x \ne 1\\
a)P = \dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x - 1 + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
b)Thay:x = 7 - 4\sqrt 3 \\
= 4 - 2.2.\sqrt 3 + 3 = {\left( {2 - \sqrt 3 } \right)^2}\\
\to P = \dfrac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + 1}}{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }} = \dfrac{{2 - \sqrt 3 + 1}}{{2 - \sqrt 3 }}\\
= \dfrac{{3 - \sqrt 3 }}{{2 - \sqrt 3 }} = 3 + \sqrt 3
\end{array}\)
