thpthang

Định lý: Nếu $T$ là tích trong trên $V$ thì $V$ có cơ sở $\left \{ v_i \right \}_{i=1}^{n}$ sao cho $T(v_i,v_j)=\delta _{ij}$ và do đó tồn tại đẳng cấu $f:\mathbb{R}^{n} \to V$ sao cho $T(f(x),f(y))=\left \langle x,y \right \rangle$.

Ánh xạ

Cho ánh xạ $f:X \to Y,$ với $A,B \subset X,$ ta có

$f\left( {A \cup B} \right) = f\left( A \right) \cup f\left( B \right)$

$f\left( {A \cap B} \right) \subset f\left( A \right) \cap f\left( B \right)$

$f\left( {A\backslash B} \right) \supset f\left( A \right)\backslash f\left( B \right)$

Chứng minh

a)

$y \in f\left( {A \cup B} \right) \Leftrightarrow \exists x \in A \cup B:f\left( x \right) = y \Leftrightarrow \left[ \begin{gathered}
\exists x \in A:f\left( x \right) = y \\
\exists x \in B:f\left( x \right) = y \\
\end{gathered} \right. \Leftrightarrow y \in f\left( A \right) \cup f\left( B \right)$

b)

$y \in f\left( {A \cap B} \right) \Rightarrow \exists x \in A \cap B:f\left( x \right) = y \Rightarrow \left\{ \begin{gathered}
\exists x \in A:f\left( x \right) = y \\
\exists x \in B:f\left( x \right) = y \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
y \in f\left( A \right) \\
y \in f\left( B \right) \\
\end{gathered} \right. \Rightarrow y \in f\left( A \right) \cap f\left( B \right)$

Xét ánh xạ $f:\mathbb{R} \to \mathbb{R}_ + ^*{\text{ }},{\text{ }}y = f\left( x \right) = {x^2},{\text{ }}A = \left\{ 1 \right\},{\text{ }}B = \left\{ { - 1} \right\},$ ta có

$f\left( A \right) = \left\{ 1 \right\},{\text{ }}f\left( B \right) = \left\{ 1 \right\} \Rightarrow 1 \in f\left( A \right) \cap f\left( B \right)$ nhưng $1 \notin f\left( {A \cap B} \right) = f\left( \emptyset \right)$

c)

$y \in f\left( A \right)\backslash f\left( B \right) \Rightarrow \left\{ \begin{gathered}
y \in f\left( A \right) \\
y \notin f\left( B \right) \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
\exists x \in A:f\left( x \right) = y \in f\left( A \right) \\
\exists x \in B:f\left( x \right) = y \in f\left( B \right) \\
\end{gathered} \right. \Rightarrow \exists x \in A\backslash B:f\left( x \right) = y \in f\left( {A\backslash B} \right)$

Xét ánh xạ $f:\mathbb{R} \to \mathbb{R}_ + ^*{\text{ }},{\text{ }}y = f\left( x \right) = {x^2},{\text{ }}A = \left\{ 1 \right\},{\text{ }}B = \left\{ { - 1} \right\},$ ta có

$f\left( A \right) = \left\{ 1 \right\},{\text{ }}f\left( B \right) = \left\{ 1 \right\} \Rightarrow 1 \in f\left( {A\backslash B} \right) = f\left( A \right)$ nhưng $1 \notin f\left( A \right)\backslash f\left( B \right) = \emptyset $

Biến đổi thế này.

Phần điều kiện bạn tự đặt nha!


$\sqrt {x + \sqrt {x - \frac{1}{4}} } + \sqrt {x - \sqrt {x - \frac{1}{4}} } = 1$


$ \Leftrightarrow 2x + 2\sqrt {{x^2} - x + \frac{1}{4}} = 1$

$ \Leftrightarrow 2x + 2\sqrt {{{\left( {x - \frac{1}{2}} \right)}^2}} = 1$

$ \Leftrightarrow 2x + 2\left| {x - \frac{1}{2}} \right| = 1$

$ \Leftrightarrow 2\left| {x - \frac{1}{2}} \right| = 1 - 2x$

Đến đây là dạng


$\left| A \right| = B \Leftrightarrow \left\{ \begin{gathered}
B \geqslant 0 \\
\left[ \begin{gathered}
A = B \\
A = - B \\
\end{gathered} \right. \\
\end{gathered} \right.$