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1: A={x∈R|x<=2}
=>A=(-∞;2]
B={x∈R|x>5}
=>B=(5;+∞)
A\(\cap\) B=(-∞;2]\(\cap\) (5;+∞)
=∅
A\(\cup\) B=(-∞;2]\(\cup\) (5;+∞)
A\B=(-∞;2]\(5;+∞)
=(-∞;2]
B\A=(5;+∞)\(-∞;2]
=(5;+∞)
\(C_{R}A\) =R\A=R\(-∞;2]
=(2;+∞)
\(C_{R}B=\) R\B=R\(5;+∞)
=(-∞;5]
2: A={x∈R|x<0 hoặc x>=2}
=>A=(-∞;0)\(\cup\) [2;+∞)
B={x∈R|-4<=x<3}
=>B=[-4;3)
A\(\cap\) B=((-∞;0)\(\cup\) [2;+∞))\(\cap\) [-4;3)
=[-4;0)\(\cup\) [2;3)
A\(\cup\) B=((-∞;0)\(\cup\) [2;+∞))\(\cup\) [-4;3)
=(-∞;+∞)
A\B=((-∞;0)\(\cup\) [2;+∞))\[-4;3)
=(-∞;-4)\(\cup\) [3;+∞)
B\A=[-4;3)\((-∞;0)\(\cup\) [2;+∞))
=[0;2)
\(C_{R}A\) =R\A=[0;2)
\(C_{R}B\) =R\B=R\[-4;3)
=(-∞;-4)\(\cup\) [3;+∞)
3: |x-1|<2
=>-2<x-1<2
=>-1<x<3
=>A=(-1;3)
|x+1|<3
=>-3<x+1<3
=>-4<x<2
=>B=(-4;2)
A\(\cap\) B=(-1;3)\(\cap\) (-4;2)
=(-1;2)
A\(\cup\) B=(-1;3)\(\cup\) (-4;2)
=(-4;3)
A\B=(-1;3)\(-4;2)
=[2;3)
B\A=(-4;2)\(-1;3)
=(-4;-1]
\(C_{R}A\) =R\A=R\(-1;3)
=(-∞;-1]\(\cup\) [3;+∞)
\(C_{R}B\) =R\B=R\(-4;2)
=(-∞;-4]\(\cup\) [2;+∞)
a: A=(-7/4; -1/2]
\(B=\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\)
\(C=\left(\dfrac{2}{3};+\infty\right)\)
b: \(\left(A\cap B\right)\cap C=\varnothing\)
\(\left(A\cup C\right)\cap\left(B\A\right)\)
\(=(-\dfrac{7}{4};-\dfrac{1}{2}]\cup\left(\dfrac{2}{3};+\infty\right)\cap\left[\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\right]\)
\(=\left(4;\dfrac{9}{2}\right)\)
\(\frac{1}{\left|x-2\right|}>2\Rightarrow\left|x-2\right|< \frac{1}{2}\Rightarrow-\frac{1}{2}< x-2< \frac{1}{2}\)
\(\Rightarrow\frac{3}{2}< x< \frac{5}{2}\)
\(\Rightarrow A=\left(\frac{3}{2};\frac{5}{2}\right)\)
\(\left|x-1\right|< 1\Rightarrow-1< x-1< 1\Rightarrow0< x< 2\)
\(\Rightarrow B=\left(0;2\right)\)
\(\Rightarrow A\cup B=\left(0;\frac{5}{2}\right)\)
\(A\backslash B=[2;\frac{5}{2})\)
Bài 3:
a: \(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)
b: \(\left(-\dfrac{11}{2};7\right)\cup\left(-2;\dfrac{27}{2}\right)=\left(-\dfrac{11}{2};\dfrac{27}{2}\right)\)
c: \(\left(0;12\right)\text{\[}5;+\infty)=\left(0;5\right)\)
d: \(R\[ -1;1)=\left(-\infty;-1\right)\cup[1;+\infty)\)
Mấy cái dấu "=" anh tự xét.
Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)
b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)