1 That man is so tall that he can almost touch the ceiling

2 The coffee is so strong that I can't go to sleep

3 My mother was so busy that she couldn't do the housework

4 The shirt is so small that I can't wear it

5 My house is so large that it can hold 100 people

6 She told such interesting stories that they all like her

7 I have such a difficult exercise that I couldn't do it

8 They pictures are so beautiful that everybody will want one

9 Day is so hot that we decided to stay indoors

10 The speaker gave such a long talk that most of the audience felt sleepy

1. That man is so tall that he can almost touch the ceiling
2. The coffee is so strong that I can't go to sleep
3. My mother was so busy that she couldn't do the housework
4. The shirt is so small that I can't wear it.
5. My house is so large that It can hold 100 people.

Ta có:

\(\left|x-5\right|+\left|2-4x\right|=\left|x-5\right|+\left|\dfrac{1}{2}-x\right|+3\left|\dfrac{1}{2}-x\right|\)

Mà \(\left|x-5\right|+\left|\dfrac{1}{2}-x\right|\ge\left|x-5+\dfrac{1}{2}-x\right|=\dfrac{9}{2}>4\)

\(\Rightarrow\left|x-5\right|+\left|2-4x\right|\ge4+3\left|\dfrac{1}{2}-x\right|>4>3\)

Vậy pt đã cho vô nghiệm

a: Trên cùng một nửa mặt phẳng bờ chứa tia Ox, ta có: \(\widehat{xOt}< \widehat{xOy}\left(50^0< 100^0\right)\)

nên tia Ot nằm giữa hai tia Ox và Oy

=>\(\widehat{xOt}+\widehat{tOy}=\widehat{xOy}\)

=>\(\widehat{tOy}=100^0-50^0=50^0\)

b: 

Vì tia Ot nằm giữa hai tia Ox và Oy

và \(\widehat{xOt}=\widehat{yOt}\left(=50^0\right)\)

nên Ot là phân giác của góc xOy
 

Tôi gửi đến bạn
Một lời cảm ơn
Hãy bỏ qua đi
Những lúc giận hờn
Tình bạn chúng ta
Hãy mãi gắn bó
Càng giận ta càng
Thân thiết nhau hơn

Mình trả lời bên dưới rồi bạn tham khảo nhé!

Ta có: p + n + e = 60 Mà số p = số e

\(\Rightarrow\) 2p + n = 60 (1)

Ta lại có: 2p = 2n \(\Leftrightarrow\) 2p - 2n = 0 (2)

Từ (1) và (2) ta có hệ phương trình sau:

\(\left\{{}\begin{matrix}2p+n=60\\2p-2n=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=e=20\\n=20\end{matrix}\right.\)

Vậy....

\(#NqHahh\)

\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\dfrac{2}{5}\right)\cdot\left(-\dfrac{5}{12}\right)^2}=\dfrac{\dfrac{2^3}{3^3}\cdot\dfrac{3^2}{4^2}\cdot\left(-1\right)}{\dfrac{2}{5}\cdot\dfrac{25}{144}}\)

\(=\dfrac{\dfrac{1}{2\cdot3}\cdot\left(-1\right)}{\dfrac{5}{72}}=-\dfrac{1}{6}\cdot\dfrac{72}{5}=-\dfrac{12}{5}\)