\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

\(\Leftrightarrow\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=2010\)

V...\(S=\left\{2010\right\}\)

^^

\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

\(\Leftrightarrow\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)

\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=2010\)

a) 5x.(x+3/4) = 0

=> x = 0

x+3/4 = 0 => x = -3/4

b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)

\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)

\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)

\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)

\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)

\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

=> x + 2017 = 0

x = -2017

a) để 2x - 3 > 0

=> 2x > 3

x > 3/2

b) 13-5x < 0

=> 5x < 13

x < 13/5

c) \(\frac{x+3}{2x-1}>0\)

=> x + 3 > 0

x > -3

d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)

Để x+7/x+3 < 1

=> 1 + 4/x+3 < 1

=> 4/x+3 < 0

=> không tìm được x thỏa mãn điều kiện

D=|x-2010| + |x-2011| + |x-2012|
D=|x-2010| + |x-2011| + |2012-x|
=>D>=|x-2010+2012-x| + |x-2011|
=>D>=|2| + |x-2011|=2 + |x-2011|
Dấu = xảy ra <=> (x-2010)(2012-x)>=0<=>2010<=x<=2012(1)
                           x-2011=0 => x =2011(2)
Từ 1,2 => x=2011
Vậy Bmin=2 khi x=2011
 

Bai 2: 

a: 2x-3>0

=>2x>3

=>x>3/2

b: =>13-5x<0

=>5x>13

=>x>13/5

c: =>2x-1>0 hoặc x+3<0

=>x>1/2 hoặc x<-3

d: =>(x+7-x-3)/(x+3)<0

=>x+3<0

=>x<-3

c) (x - 1)(y + 1) = 5

=> x - 1 và y + 1 là các ước của 5

Ư(5) = {1; -1; 5; -5}

Lập bảng giá trị:

Vậy các cặp (x,y) cần tìm là:

(2; 4); (6; 0); (0; -6); (-4; -2). 

(x-1).(y+1) = 5

=> x-1 và y+1 ∈ Ư(5) = {-1;-5;1;5}

Ta có bảng sau : 

vậy ta có các cặp số (x;y) là :

(0;-6);(-4;-2);(2;4);(6;0)

CÓ ai làm câu a, b ko nhỉ 

a)(3/2-0,5)/x=7/2+1/4

(3/2-1/2)/x=14/4+1/4

1/x=15/4

x=1:15/4

x=4/15

b)(x*0,25+2010)*2011=(53+2010)*(2012-1)

(x*0,25+2010)*2011=2063*2011

=>0,25x+2010=2063

0,25x=2063-2010

0,25x=53

x=53/0,25

x=212

\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)

\(\left(\frac{x+10}{2008}+1\right)+\left(\frac{x+9}{2009}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

mà \(\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)\ne0\)

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=-2018\)

Vậy,.............

Ta có: \(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)

\(\Rightarrow\frac{x+10}{2008}+1+\frac{x+9}{2009}+1=\frac{x+8}{2010}+1+\frac{x+7}{2011}+1\)

\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)

\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(\Rightarrow x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

Do \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=-2018\)

Vậy \(x=-2018\)

\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}.\)

\(\Rightarrow\frac{x+10}{2008}+\frac{x+9}{2009}-\frac{x+8}{2010}-\frac{x+7}{2011}=0\)

\(\frac{x+10}{2008}+1+\frac{x+9}{2009}+1-\frac{x+8}{2010}-1-\frac{x+7}{2011}-1=0\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(\left(x+2018\right).\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

mà \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)

=> x + 2018 = 0

x = -2018

Bài 1 :

7^2x+3 . 7^5-2x : 7^x + 7^x = 1

- > 7^{(2x+3)+(5-2x)-7} + 7^x = 1

- > 7¹ + 7^x = 1

- > 7^x = 1 - 7 = -6 - > x thuộc rỗng