Đề bài sai, đề đúng phải là: \(\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{abc+ca+c}=1\)

Phản ví dụ chứng minh đề bài sai: lấy \(a=1;b=2;c=\dfrac{1}{2}\) thỏa mãn \(abc=1\)

Khi đó thay vào biểu thức:

\(\dfrac{1}{1.2+1+1}+\dfrac{1}{2.\dfrac{1}{2}+2+1}+\dfrac{1}{1.2.\dfrac{1}{2}+2.\dfrac{1}{2}+2}=\dfrac{3}{4}\ne1\)

nhanh nhất là vận động viên thứ 1

chậm nhất là vận động viên thứ 2

ĐK: \(x\ge0;x\ne4;x\ne9\)

\(Q=\left(\dfrac{x-3\sqrt{x}}{9-x}+1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)

\(=\left[-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+1\right]:\left[\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right]\)

\(=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}.\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-2}\)

\(=\dfrac{3}{2-\sqrt{x}}\)

Giúp e vs

Ta có: Ox//CD(gt)

\(\Rightarrow\widehat{OCD}+\widehat{COx}=180^0\)( 2 góc trong cùng phía)

\(\Rightarrow\widehat{COx}=180^0-\widehat{OCD}=180^0-120^0=60^0\)

\(\Rightarrow\widehat{ACx}=\widehat{AOC}-\widehat{COx}=110^0-60^0=50^0\)

Ta có: \(\widehat{ACx}+\widehat{OAB}=50^0+130^0=180^0\)

Mà 2 góc này là 2 góc trong cùng phía

=> AB//Cx//CD

Bài 1:

Vì \(Ox//CD\) nên \(\text{∠}OCD+\text{∠}COx=180\) ( góc trong cũng phía )

\(\Rightarrow\text{∠}COx=180-120=60\)

Mặt khác, \(\text{∠}COx+\text{∠}xOA=\text{∠}AOC\)

\(\Rightarrow\text{∠}xOA=110-60=50\)

Ta có \(\text{∠}xOA+\text{∠}OAB=50+130=180\) mà hai góc này lại ở vị trí góc trong cùng phía 

\(\Rightarrow Ox//AB\) \(\Rightarrow AB//CD\) ( cùng // Ox )

g: =>x-1-2x+1-9+x=0

=>-9=0(vô lý)

h: \(\Leftrightarrow x^2+x-12-6x+4-x^2+8x-16=0\)

=>3x-24=0

hay x=8

i: \(\Leftrightarrow x^3+6x^2+9x-3x-x^3-6x^2-12x-8-1=0\)

=>-6x-9=0

=>6x=-9

hay x=-3/2

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PTHH: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)

                0,5<-----------0,25------->0,25------------->0,25

\(\rightarrow C_{M\left(CH_3COOH\right)}=\dfrac{0,5}{0,05}=10M\)

\(m_{dd}=100+6-0,25.2=105,5\left(g\right)\\ \rightarrow C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{0,25.142}{105,5}.100\%=33,65\%\)

2. In the future, government organizations will easily access information on RFID chips.

3. I think they will introduce more effective methods of surveillance.

4. Travelling will become easier and we won't show any ID within the E.U.

5. The police will use state-of-the-art technology to improve personal security.

11 A

12 chắc hẳn thiếu remember B

13 C

14 D

15 A

16 B

17 C

18 C

19 D

20 A

21 C

22 D

15 B

16 A

17 A

18 A

19 C

20 A