KIỂM TRA GIẢI TÍCH 12 – HOÀNG LÊ KHA – BÀI KIỂM SỐ 1

Gợi ý:

ĐK: \[x > \frac{3}{2}\]

\[\begin{array}{l}
{\rm{lo}}{{\rm{g}}_2}x.{\log _3}\left( {2x – 3} \right) = 2{\log _2}x\\
\Leftrightarrow {\rm{lo}}{{\rm{g}}_2}x.{\log _3}\left( {2x – 3} \right) – 2{\log _2}x = 0\\
\Leftrightarrow {\rm{lo}}{{\rm{g}}_2}x\left[ {{{\log }_3}\left( {2x – 3} \right) – 2} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\rm{lo}}{{\rm{g}}_2}x = 0\\
{\log _3}\left( {2x – 3} \right) – 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {loai} \right)\\
x = 6
\end{array} \right.
\end{array}\]

Gợi ý:

A. sai: \[y’ = \frac{1}{{2x\ln 2}}\]

B. sai: Tập giá trị thuộc R

C. Đúng

D. Sai

SGK NC/108

Gợi ý:

\[\begin{array}{l}
{3^{1 + x}} + {3^{1 – x}} = 10\\
\Leftrightarrow {3.3^x} + \frac{3}{{{3^x}}} – 10 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 1
\end{array} \right.
\end{array}\]

Gợi ý:

\[y’ + 1 = 0 \Leftrightarrow \frac{{2x}}{{{x^2} + 1}} + 1 \Leftrightarrow x = – 1\]

Gợi ý:

\[\begin{array}{l}
T = M{\left( {1 + r\% } \right)^n}\\
2M = M{\left( {1 + 1,075} \right)^n} \Leftrightarrow n = {\log _{1,075}}2 \approx 9,6
\end{array}\]

Gợi ý:

ĐK: \[{3.2^x} – 1 > 0 \Leftrightarrow {2^x} > \frac{1}{3} \Leftrightarrow x > – {\log _2}3\]

\[\begin{array}{l}
{\log _4}\left( {{{3.2}^x} – 1} \right) = x – 1\\
\Leftrightarrow {3.2^x} – 1 = {4^{x – 1}}\\
\Leftrightarrow {3.2^x} – \frac{1}{4}{2^{2x}} – 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{2^x} = 6 + 4\sqrt 2 \\
{2^x} = 6 – 4\sqrt 2
\end{array} \right.\\
\left[ \begin{array}{l}
{x_1} = {\log _2}\left( {6 + 4\sqrt 2 } \right)\\
{x_2} = {\log _2}\left( {6 – 4\sqrt 2 } \right)
\end{array} \right.\\
{x_1} + {x_2} = {\log _2}\left[ {\left( {6 + 4\sqrt 2 } \right)\left( {6 – 4\sqrt 2 } \right)} \right] = 2
\end{array}\]

Gợi ý:

\[\log _3^2x + \sqrt {\log _3^2x + 1} – 2m – 1 = 0\]

\[t = {\log _3}x;\,\,t \in \left[ {0;\sqrt 3 } \right]\]

\[\frac{{{t^2} + \sqrt {{t^2} + 1} – 1}}{2} = m\]

\[g\left( t \right) = \frac{{{t^2} + \sqrt {{t^2} + 1} – 1}}{2}\]

\[g’\left( t \right) = t + \frac{t}{{2\sqrt {{t^2} + 1} }};g’\left( t \right) = 0 \Leftrightarrow t = 0\]

\[g\left( 0 \right) = 0;g\left( {\sqrt 3 } \right) = 2\,\,\, \Rightarrow m \in \left[ {0;2} \right]\]

Gợi ý:

\[ – {x^2} + 5x – 6 > 0 \Leftrightarrow 2 < x < 3\]

Gợi ý:

\[{\left( {\sqrt 6 – \sqrt 5 } \right)^x} > \sqrt 6 + \sqrt 5 \]

\[{\left( {\sqrt 6 – \sqrt 5 } \right)^x} = \frac{1}{{{{\left( {\sqrt 6 + \sqrt 5 } \right)}^x}}} = {\left( {\sqrt 6 + \sqrt 5 } \right)^{ – x}} > {\left( {\sqrt 6 + \sqrt 5 } \right)^1}\]

\[x < – 1\]

Gợi ý:

\[\begin{gathered}
\sqrt {{{\left( {{a^\pi } + {b^\pi }} \right)}^2} – {{\left( {{4^{\frac{1}{\pi }}}ab} \right)}^\pi }} \,\, \hfill \\
\sqrt {{a^{2\pi }} + 2{a^\pi }{b^\pi } + {b^{2\pi }} – 4{a^\pi }{b^\pi }} \,\, \hfill \\
\sqrt {{a^{2\pi }} – 2{a^\pi }{b^\pi } + {b^{2\pi }}} \,\, \hfill \\
\sqrt {{{\left( {{a^\pi } – {b^\pi }} \right)}^2}} = \left| {{a^\pi } – {b^\pi }} \right| = {b^\pi } – {a^\pi } \hfill \\
\end{gathered} \]

Gợi ý:

\[x – 1 \ne 0 \Leftrightarrow x \ne 1\]

SGK CB/57

Gợi ý:

Đk: \[x > 0\]

\[x = {2^{\frac{1}{2}}} = \sqrt 2 \]

Gợi ý:

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