2. B

3. A

4. A

5. D

6. C

7. B

8. D

9. C

10. A

2 B

3 A

4 A

5 D

6 C

7 B

8 D

9 C

10 A

B nhá bạn

🧠

CHỊU

|x-1|+|2x-2|+|3x-3|=12

=>\(\left|x-1\right|+2\left|x-1\right|+3\left|x-1\right|=12\)

=>\(6\left|x-1\right|=12\)

=>|x-1|=2

=>\(\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

cảm ơn

1.y tá

2.bác sĩ thú y

3.thợ làm tóc

4.cảnh sát

5.nhân viên bán hàng

Nurse : y tá

Vet : bác sĩ thú y

Hairdresser: thợ cắt tóc

Policeman: cảnh sát

Shop assistand: nhân viên bán hàng

                                      ~ THE END~

1.cháu trai ngoại/nội

2.cháu gái ngoại/nội

3.người trưởng thành

4.cha mẹ

5.ông bà ngoại/nội

6.con trai

7.hàng xóm

8.vợ

9.chồng

10.hải tặc

1.cháu trai

2.cháu gái

3.lớn lên

4.bố mẹ

5.ông bà

6.con trai

7.hàng xóm

8.vợ

9.chồng

10.cướp biển

a) \(\dfrac{x+2004}{x+2005}+\dfrac{x+2005}{2006}< \dfrac{x+2006}{2007}+\dfrac{x+2007}{2008}\\ \Rightarrow\left(\dfrac{x+2004}{2005}-1\right)+\left(\dfrac{x+2005}{2006}-1\right)< \left(\dfrac{x+2006}{2007}-1\right)+\left(\dfrac{x+2007}{2008}-1\right)\\ \Rightarrow\dfrac{x-1}{2005}+\dfrac{x-1}{2006}< \dfrac{x-1}{2007}+\dfrac{x-1}{2008}\\ \Rightarrow\dfrac{x-1}{2005}+\dfrac{x-1}{2006}-\dfrac{x-1}{2007}-\dfrac{x-1}{2008}< 0\\ \)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}-\dfrac{1}{2008}\right)< 0\left(a\right)\)

Nhận thấy: \(\dfrac{1}{2005}>\dfrac{1}{2007},\dfrac{1}{2006}>\dfrac{1}{2008}\\ \Rightarrow\dfrac{1}{2005}-\dfrac{1}{2007}>0,\dfrac{1}{2006}-\dfrac{1}{2008}>0\\ \Rightarrow\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}-\dfrac{1}{2008}>0\)

\(\left(a\right)\Rightarrow x-1< 0\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

b) \(\dfrac{x-2}{2002}+\dfrac{x-4}{2000}< \dfrac{x-3}{2001}+\dfrac{x-5}{1999}\\ \Rightarrow\left(\dfrac{x-2}{2002}-1\right)+\left(\dfrac{x-4}{2000}-1\right)< \left(\dfrac{x-3}{2001}-1\right)+\left(\dfrac{x-5}{1999}-1\right)\\ \Rightarrow\dfrac{x-2004}{2002}+\dfrac{x-2004}{2000}< \dfrac{x-2004}{2001}+\dfrac{x-2004}{1999}\\ \Rightarrow\dfrac{x-2004}{2002}+\dfrac{x-2004}{2000}-\dfrac{x-2004}{2001}-\dfrac{x-2004}{1999}< 0\\ \)

\(\Rightarrow\left(x-2004\right)\left(\dfrac{1}{2002}+\dfrac{1}{2000}-\dfrac{1}{2001}-\dfrac{1}{1999}\right)< 0\left(b\right)\)

Nhận thấy: \(\dfrac{1}{2002}< \dfrac{1}{2001},\dfrac{1}{2000}< \dfrac{1}{1999}\Rightarrow\dfrac{1}{2002}-\dfrac{1}{2001}< 0,\dfrac{1}{2000}-\dfrac{1}{1999}< 0\\ \Rightarrow\dfrac{1}{2002}+\dfrac{1}{2000}-\dfrac{1}{2001}-\dfrac{1}{1999}< 0\)

\(\left(b\right)\Rightarrow x-2004>0\Leftrightarrow x>2004\)