Em nên chia bài này ra làm 2, 3 nhé vì liền một lúc 4 đoạn như vậy mng sẽ nản hoặc khó để làm em nhé!

5T

6F

7T

8F

F - F - T - F

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.

1-wool

2-cotton

3-metal

4-letter

5-glass(cs thể th vì nó thừa)

6-wood 

Mình nghĩ nó thế này:

2. cotton

4.leather

5.wool

6.wood

Còn 1 với 3 mình không biết

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