Đăng từng bài thôi bn 

Ok bạn

\(cosC=\dfrac{7^2+3^2-7^2}{2\cdot7\cdot3}=\dfrac{3}{14}\)

=>góc C=78 độ

\(m_b=\sqrt{\dfrac{7^2+3^2}{2}-\dfrac{7^2}{4}}=\dfrac{\sqrt{67}}{2}\)

Theo thứ tự các câu từ trên xuống:

\(\left\{b;d\right\}\)

\(\left\{a;b;c;d;e;f\right\}\)

\(\left\{a;c\right\}\)

\(\left\{e;f\right\}\)

\(\left\{b;d\right\}\)

\(\left\{a;b;c;d;e;f\right\}\)

\(\varnothing\)

\(X=\left\{-3;-2;-1;0;1;2;3\right\}\)

\(Y=\left\{0;1;2;3;4\right\}\)

Làm biếng ghi các kí tự nên theo thứ tự từ trên xuống là các câu a-b-c...

\(\left\{0;1;2;3\right\}\)

\(\left\{-3;-2;-1;0;1;2;3;4\right\}\)

\(\left\{-3;-2;-1\right\}\)

\(\left\{4\right\}\)

\(\left\{0;1;2;3\right\}\)

\(\left\{-3;-2;-1;0;1;2;3;4\right\}\)

\(\varnothing\)

P/s: Cái này bạn vẽ trục số ra sẽ rõ nhé!

      `X=(-oo;5]` và `Y=(-2;+oo)`

`a)X nn Y=(-2;5]`

`b)X uu Y= RR`

`c)X \\ Y=(-oo;2]`

`d)Y \\ X=(5;+oo)`

`e)(X nn Y) nn (X uu Y)= (-2;5]`

`f)(X uu Y) uu (X uu Y)=RR`

\(X=[-4;3)\)

\(Y=(-1;6]\)

\(X\cap Y=\left(-1;3\right)\)

\(X\cup Y=\left[-4;6\right]\)

\(X\backslash Y=\left[-4;-1\right]\)

\(Y\backslash X=\left[3;6\right]\)

\(\left(A\cap B\right)\cup C=[0;\sqrt{5})\)

\(\left(A\cup B\right)\cap C=[0;1)\cup[2;\sqrt{5})\)

\(A\cap\left(B\cup C\right)=[0;1)\)

$\left(A\cap B\right)\cup C=[0;\sqrt{5})$

$\left(A\cup B\right)\cap C=[0;1)\cup [2;\sqrt{5})$

$A\cap \left(B\cup C\right)=[0;1)$

Chứng minh các biểu thức khồn phụ thuộc vào x

\(E=\left(cos^6x+sin^6x\right)+sin^4x\left(sin^2x+cos^2x\right)+4sin^2x.cos^2x-sin^2x\)

\(=\left(cos^2x+sin^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+sin^4x+4sin^2x.cos^2x-sin^2x\)

\(=1+sin^2x.cos^2x+sin^4x-sin^2x\)

\(=cos^2x+sin^2x\left(sin^2x+cos^2x\right)\)

\(=cos^2x+sin^2x=1\)

2.

\(E=\left\{-3;-2;-1;0;1;2;3\right\}\)

\(A=\left\{-2;0;2\right\}\)

\(B=\left\{-1;0;1;2;3\right\}\)

\(A\cap B=\left\{0;2\right\}\)

\(A\cup B=\left\{-2;-1;0;1;2;3\right\}\)

\(C_EA=\left\{-3;-1;1;3\right\}\)

\(C_EB=\left\{-3;-2\right\}\)

3.

\(A=\left\{0;1;2;3;4\right\}\)

\(B=\left\{1;3;9;27\right\}\)

\(A\cap B=\left\{1\right\}\)

\(A\cup B=\left\{0;1;2;3;4;9;27\right\}\)

\(A\backslash B=\left\{0;2;3;4\right\}\)

\(B\backslash A=\left\{3;9;27\right\}\)