a)\(\frac{x-10}{2010}\)+ \(\frac{x-3}{2003}\)+\(\frac{x-2}{2002}\)= -3

=> \(\frac{x-10}{2010}\)+1+ \(\frac{x-3}{2003}\)+ 1+\(\frac{x-2}{2002}\)+1= -3 +1 + 1 + 1

=> \(\frac{x-10+2010}{2010}\)+ \(\frac{x-3+2003}{2003}\)+\(\frac{x-2+2002}{2002}\)= 0

=>\(\frac{x+2000}{2010}\)+ \(\frac{x+2000}{2003}\)+\(\frac{x+2000}{2002}\)= 0

=>(x + 2000)(\(\frac{1}{2010}\)+ \(\frac{1}{2003}\)+\(\frac{1}{2002}\)) = 0

=> x + 2000 = 0

hoặc

=>\(\frac{1}{2010}\)+ \(\frac{1}{2003}\)+\(\frac{1}{2002}\)= 0

Mà : \(\frac{1}{2010}\)> 0

\(\frac{1}{2003}\)> 0

\(\frac{1}{2002}\)> 0

Cộng vế theo vế của các bất đẳng thức trên , ta có:

\(\frac{1}{2010}\)+\(\frac{1}{2003}\)+\(\frac{1}{2002}\)>0

=> x + 2000 = 0

=> x = 0 -2000 = -2000

Vậy x = -2000

Nhường các bạn câu 2 :(

de sai ban

   3³ - 2x = 7¹⁰⁰⁵ . 7¹⁰¹³ . 7²⁰¹⁷

   3³ - 2x = 7⁴⁰³⁵

          2x = 3³ - 7⁴⁰³⁵

               x = ( 3³ - 7⁴⁰³⁵ ) : 2

135 - 3 ( x + 1 ) = 30

3 ( x + 1 ) = 105

x + 1 = 35

x = 34

\(1005^2\cdot1005^x=1005^7\)

\(1005^{2+x}=1005^7\)

=> x + 2 = 7

=> x = 5

Vậy,........

\(135-3.\left(x+1\right)=30\)

\(\Rightarrow3.\left(x+1\right)=105\)

\(\Rightarrow x+1=35\)

\(\Rightarrow x=34\)

\(1005^2.1005^x=1005^7\)

\(\Rightarrow1005^x=1005^7\div1005^2\)

\(\Rightarrow1005^x=1005^{7-2}\)

\(\Rightarrow1005^x=1005^2\)

\(\Rightarrow x=2\)

\(1005^2\cdot1005^x=1005^7\)

\(1005^{x+2}=1005^7\)

\(\Rightarrow x+2=7\)

\(\Leftrightarrow x=5\)

Tra loi

X = 5

Hok tot 

Bn cs the chep , tham khao bn Bui Huyen

\(1005^2.1005^x=1005^7\)

\(\Rightarrow1005^x=1005^7:1005^2\)

\(\Rightarrow1005^x=1005^5\)

\(\Rightarrow x=5\)

Vậy x = 5

x = 4

x= 5

\(3^{x-1}=27\)

\(3^{x-1}=3^3\)

\(=>x-1=3\)

\(=>x=4\)

vậy \(x=4\)

\(1005^2.1005^x=1005^7\)

\(=>2x=7\)

\(=>x=3,5\)

vậy \(x=3,5\)

a)3^x-1=27

3^x-1=3³

=>x-1=3

xx=3+1=4

b) 1005² .1005^x=1005⁷

1005^x=1005⁷-1005²

1005^x=1005⁵

=>x= 5

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\frac{1}{3}-\frac{1}{x+2}=\frac{2010}{2011}\)

\(\frac{1}{x+2}=\frac{1}{3}-\frac{2010}{2011}\)

\(\frac{1}{x+2}=\frac{1}{2011}\)

\(\Rightarrow x+2=2011\)

\(x=2009\)

Đặt biểu thứ là A

2A=2/1.3+2/2.5+...+2/x.x+2

2A=1-1/3+1/3-1/5+.......+1/x-1/x+2

2A=1-1/x+2

x=2009 nha ban

chuc ban hoc gioi

tk cho minh nha

Bài 1 : 

1 + 4 + 7 + .......... + 79

Số các số hạng của dãy số trên là : ( 79 - 1 ) : 3 + 1 = 27 ( số )

Tổng các số hạng là : ( 79 + 1 ) . 27 : 2 = 1080

70 . 939 + 61 . 135  - 61 . 65

= 70 . 939 + 61 . ( 135 - 65 )

= 70 .939 + 61 . 70

= 70 . ( 939 + 61 )

= 70 . 1000 = 70000