a) Ta có:

\(\begin{array}{l}\left. \begin{array}{l}M \in \left( R \right)\\MH \bot \left( P \right)\\\left( R \right) \bot \left( P \right)\end{array} \right\} \Rightarrow MH \subset \left( R \right)\\\left. \begin{array}{l}M \in \left( R \right)\\MK \bot \left( Q \right)\\\left( R \right) \bot \left( Q \right)\end{array} \right\} \Rightarrow MK \subset \left( R \right)\end{array}\)

b) Ta có:

\(\left. \begin{array}{l}MH \bot \left( P \right) \Rightarrow MH \bot a\\MK \bot \left( Q \right) \Rightarrow MK \bot a\\MH,MK \subset \left( R \right)\end{array} \right\} \Rightarrow a \bot \left( R \right)\)

a) (\(-\infty;0\)] \(\cup\left[1;2\right]\cup\) [\(3;+\infty\))

b) (\(-\infty;4\)] \(\cup\) [\(5;+\infty\))

c) \(\left(-2;1\right)\cup\left(3;7\right)\)

d) (\(-1;1\)] \(\cup\) [\(4;5\))

Ta có:

\(\left. \begin{array}{l}\left( Q \right)\parallel \left( R \right)\\\left( P \right) \cap \left( Q \right) = a'\\\left( P \right) \cap \left( R \right) = b'\end{array} \right\} \Rightarrow a'\parallel b'\)

Vậy nếu \(\left( Q \right)\parallel \left( R \right)\) thì \(a'\parallel b'\); nếu \(\left( Q \right) \equiv \left( R \right)\) thì \(a' \equiv b'\).

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

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

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

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

Ta chứng minh: \(2\left(a^2-ab+b^2\right)^2\ge b^4+a^4\left(1\right)\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)^2\ge0\)( Luôn đúng \(\forall a;b\))

Tương tự có: \(2\left(b^2-bc+c^2\right)^2\ge b^4+c^4\left(2\right)\)

Và: \(2\left(c^2-ca+a^2\right)^2\ge a^4+c^4\left(3\right)\)

Ta nhân các vế trên ta được: \(8\left(a^2-ab+b^2\right)^2\left(b^2-bc+c^2\right)^2\left(c^2-ca+a^2\right)^2\ge\left(a^4+b^4\right)\left(b^4+c^4\right)\left(c^4+a^4\right)=8\)

Hay: \(\left(a^2-ab+b^2\right)\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\ge1\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

Trâu bò:

Giả sử c = min{a,b,c}

Đặt a =x +c; b = y +c;c=c thì x,y >= 0

C/m: \(8\left[\left(a^2-ab+b^2\right)\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\right]^2\ge\left(a^4+b^4\right)\left(b^4+c^4\right)\left(c^4+a^4\right)\)

Xét hiệu hai vế thu được:

\(c*(12*x^3*y^8-8*x^4*y^7+16*x^5*y^6+16*x^6*y^5-8*x^7*y^4+12*x^8*y^3)+c^2*(18*x^2*y^8-16*x^3*y^7+60*x^4*y^6+60*x^6*y^4-16*x^7*y^3+18*x^8*y^2)+c^3*(12*x*y^8+16*x^2*y^7+88*x^4*y^5+88*x^5*y^4+16*x^7*y^2+12*x^8*y)+c^4*(6*y^8+16*x*y^7+32*x^2*y^6-32*x^3*y^5+242*x^4*y^4-32*x^5*y^3+32*x^6*y^2+16*x^7*y+6*x^8)+7*x^4*y^8+c^5*(16*y^7+16*x*y^6+88*x^3*y^4+88*x^4*y^3+16*x^6*y+16*x^7)-16*x^5*y^7+c^6*(24*y^6-16*x*y^5+60*x^2*y^4+60*x^4*y^2-16*x^5*y+24*x^6)+24*x^6*y^6+c^7*(16*y^5-8*x*y^4+16*x^2*y^3+16*x^3*y^2-8*x^4*y+16*x^5)-16*x^7*y^5+c^8*(8*y^4-16*x*y^3+24*x^2*y^2-16*x^3*y+8*x^4)+7*x^8*y^4\)Dấu " * " là nhân.

Dễ thấy nó đúng -> qed

a) (0, 7)

b) (2, 5)

c) [3, +∞)

a)(0,7).

b)(2,5).

c)(3,\(+\infty\)).

@alibaba nguyễn : Giúp với ông ei :) Chắc ông cũng học đến cái này r :))

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

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