1. Mohenjo-Daro was one of the first city.

2. If we look at a map of Mohenjo-Daro, we can see the straight street.

3. Rome was the capital of the Roman Republic and Empire for a thousand years.

4. London was the biggest city in the world between 1831 and 1925.

5. Tokyo is the biggest city in the world now.

Bài 1 : 

Thay x = 2 ; y = -1/2 ta được 

\(B=-8+2.4\left(-\dfrac{1}{2}\right)-4.2.\left(\dfrac{1}{4}\right)+2\left(-\dfrac{1}{2}\right)-3\)

\(=-8-4-2-1-3=-18\)

a) Ta có: \(\left(2x-3\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};-\dfrac{4}{3}\right\}\)

b) Ta có: \(x^3-3x^2+3x-1=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2-2x+1-x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-3x\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)

Vậy: S={0;1;3}

c) Ta có: \(x^2+x=2x+2\)

\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy: S={-1;2}

d) Ta có: \(\left(x-1\right)^2=2\left(x^2-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1-2x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)Vậy: S={1;-3}

e) Ta có: \(2\left(x+2\right)^2-x^3-8=0\)

\(\Leftrightarrow2\left(x+2\right)^2-\left(x^3+8\right)=0\)

\(\Leftrightarrow2\left(x+2\right)\cdot\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x+4-x^2+2x-4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\cdot\left(-x^2+4x\right)=0\)

\(\Leftrightarrow-x\left(x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\end{matrix}\right.\)

Vậy: S={0;-2;4}

a: \(P=-\left|5-x\right|+2019\le2019\forall x\)

Dấu '=' xảy ra khi x=5

a: P(x)=2x^3-2x^3+x^2+3x^2-4x^2-3x+5x+1=-3x+6

b: P(0)=-3*0+6=6

P(-1)=6+3=9

P(1/3)=-1+6=5

c: P(x)=0

=>-3x+6=0

=>-3x=-6

=>x=2

P(x)=1

=>-3x+6=1

=>-3x=-5

=>x=5/3

used to deliver

used to be

used to go

used to drive

used to spend

used to believe

used to work

used to serve

1. used to deliver

2. used to be 

3. used to go

4. used to drive

5. used to spend

6. used to believe

7.used to work

8. used to serve