1. Đ

2. S 

3. Đ 

4. Đ

5. S

\(\left(27+11\right)\cdot\left(512-\left[14\cdot\left(64-4^2\right):2\right]\right)\\ =33\cdot\left[512-\left[14\cdot\left(64-16\right):2\right]\right]\\ =33\cdot\left(512-14\cdot48:2\right)\\ =33\cdot\left(512-14\cdot24\right)\\ =33\cdot\left(512-336\right)\\ =33\cdot176\\ =5808\)

-6∈Z; -6∈Q

22∈N; 22∈Z; 22∈Q

\(-\frac{2}{23}\in Q\)

N⊂Z; N⊂Q

Z⊂Q

\(-\frac57\notin N;-\frac57\notin Z\)

-21∉N

\(1\frac34\notin N;1\frac34\notin Z\)

\(-4\notin N\\ -\dfrac{5}{3}\in Q\\ -8\in Q\\ -\dfrac{2}{9}\notin Z\\ -\dfrac{1}{11}\notin Z\\ -\dfrac{2}{7}\in Q\\ -\dfrac{2}{19}\notin N\\ N\subset Q\)

Bài 4:

a: \(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\cdots+\left|x+\frac{1}{2019\cdot2020}\right|=2020x\)

=>2020x>=0

=>x>=0

Phương trình sẽ trở thành:

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{2019\cdot2020}=2020x\)

=>\(2020x=2019x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\right)\)

=>\(2020x=2019x+1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2019}-\frac{1}{2020}\)

=>\(2020x=2019x+1-\frac{1}{2020}=2019x+\frac{2019}{2020}\)

=>\(x=\frac{2019}{2020}\) (nhận)

b: \(\left|x+\frac{1}{1\cdot3}\right|+\left|x+\frac{1}{3\cdot5}\right|+\cdots+\left|x+\frac{1}{197\cdot199}\right|=100x\)

=>100x>=0

=>x>=0

Phương trình sẽ trở thành:

\(x+\frac{1}{1\cdot3}+x+\frac{1}{3\cdot5}+\cdots+x+\frac{1}{197\cdot199}=100x\)

=>\(99x+\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{197\cdot199}\right)=100x\)

=>\(100x=99x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{197\cdot199}\right)\)

=>\(x=\frac12\left(1-\frac13+\frac13-\frac15+\ldots+\frac{1}{197}-\frac{1}{199}\right)=\frac12\left(1-\frac{1}{199}\right)=\frac12\cdot\frac{198}{199}=\frac{99}{199}\)

c: \(\left|x+\frac12\right|+\left|x+\frac16\right|+\left|x+\frac{1}{12}\right|+\ldots+\left|x+\frac{1}{110}\right|=11x\)

=>11x>=0

=>x>=0

Phương trình sẽ trở thành:

\(x+\frac12+x+\frac16+\cdots+x+\frac{1}{110}=11x\)

=>\(11x=10x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\right)\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\) (nhận)

Bài 6:

a: \(\left|x+5\right|+\left|4-x\right|=9\)

mà \(\left|x+5\right|+\left|4-x\right|\ge\left|x+5+4-x\right|=9\)

nên (x+5)(4-x)>=0

=>(x+5)(x-4)<=0

=>-5<=x<=4

b: \(\left|x-\frac23\right|+\left|x-\frac34\right|=\frac{1}{12}\)

=>\(\left|x-\frac23\right|+\left|\frac34-x\right|=\frac{1}{12}\)

mà \(\left|x-\frac23\right|+\left|\frac34-x\right|\ge x-\frac23+\frac34-x=\frac{1}{12}\)

nên \(\left(x-\frac23\right)\left(x-\frac34\right)\le0\)

=>\(\frac23\le x\le\frac34\)

c: \(2\left|x-3\right|+\left|2x+5\right|=11\)

=>\(\left|2x-6\right|+\left|2x+5\right|=11\)

=>\(\left|2x+5\right|+\left|6-2x\right|=11\)

mà \(\left|2x+5\right|+\left|6-2x\right|\ge2x+5+6-2x=11\)

nên (2x+5)(2x-6)<=0

=>\(-\frac52\le x\le3\)

d: \(\left|x-3\right|+\left|5-x\right|-2\left|x-4\right|=2\)

=>|x-3|-2|x-4|+|x-5|=2(1)

TH1: x<3

=>x-3<0; x-4<0; x-5<0

(1) sẽ trở thành:

3-x-2(4-x)+5-x=2

=>8-2x-8+2x=2

=>0x=2(vô lý)

=>Loại

TH2: 3<=x<4

=>x-3>=0; x-4<0; x-5<0

(1) sẽ trở thành:

x-3-2(4-x)+5-x=2

=>x-3-8+2x+5-x=2

=>2x-6=2

=>2x=8

=>x=4(loại)

TH3: 4<=x<5

=>x-3>0; x-4>=0; x-5<0

(1) sẽ trở thành:

x-3-2(x-4)+5-x=2

=>x-3-2x+8+5-x=2

=>-2x+10=2

=>-2x=-8

=>x=4(nhận)

TH4: x>=5

=>x-3>0; x-4>0; x-5>=0

(1) sẽ trở thành: x-3+2(x-4)+x-5=2

=>2x-8+2x-8=2

=>4x-16=2

=>4x=18

=>x=18/4(loại)

Vậy: x=4

Bải 3:

a: \(\left|5-\frac23x\right|\ge0\forall x;\left|\frac23y-4\right|\ge0\forall y\)

Do đó: \(\left|5-\frac23x\right|+\left|\frac23y-4\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}5-\frac23x=0\\ \frac23y-4=0\end{cases}\Rightarrow\begin{cases}\frac23x=5\\ \frac23y=4\end{cases}\Rightarrow\begin{cases}x=5:\frac23=\frac{15}{2}\\ y=4:\frac23=4\cdot\frac32=6\end{cases}\)