40793.96757982241

bài này dễ 

bằng 40793.96757982241

12.(x-3):3=64+8

12.(x-3):3=72

12.(x-3)   =72.3

12.(x-3)   =216

     x-3     =216:12

      x-3    =18

      x       =18+3

       x      =21

100-7.(x-5)=65

       7.(x-5)=100-65

       7.(x-5)=35

           x-5=35:7

           x-5=5

           x   =5+5

            x   =10

Vậy x = 10

5x -201=2 x 2 x 2 x 2 x 2

5x - 201=32

5x          =32+201

5x          = 233

tự làm tiếp nhé 

x . 2 + x . 5 + x . 8 = x . 9 + x . 7 + x . 1/3 + 201

x . ( 2 + 5 + 8 ) = x . ( 9 + 7 + 1/3 ) +201

x . 15 = x . 49/3 + 201

15x = 49/3x + 201

15x - 201 = 49/3x

giảm mỗi vế đi 15x , ta có :

-201 = 4/3x

x = (-201) : 4/3

x = -150,75

Vậy x = -150,75

\(x\times2+x\times5+x\times8=x\times9+x\times7+x\times13+201\)

\(x\times\left(2+5+8\right)=x\times\left(9+7+13\right)+201\)

\(x\times15-x\times29=201\)

\(x\times\left(-14\right)=201\)

\(x=201:\left(-14\right)=\)Số dài quá bạn ơi!

Mk chưa hiểu chỗ [ - 4 ] là gì

Bài này dễ mà, you có thể tự làm được

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)

\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0

Vậy x = -2004

a) \(\frac{7}{13}x\frac{7}{15}-\frac{5}{12}x\frac{21}{39}+\frac{49}{91}x\frac{8}{15}=\frac{7}{13}x\frac{7}{15}-\frac{5}{12}x\frac{7}{13}+\frac{7}{13}x\frac{8}{15}\)

\(=\frac{7}{13}x\left(\frac{7}{15}-\frac{5}{12}+\frac{8}{15}\right)=\frac{7}{13}x\left(1-\frac{5}{12}\right)=\frac{7}{13}x\frac{7}{12}=\frac{49}{156}\)

b) \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)x0=0\)

Bài 2:

a) \(0,5+\left(x-\frac{15}{2}\right):\frac{1}{2}=\frac{9}{2}\)

\(\left(x-\frac{15}{2}\right):\frac{1}{2}=4\)

\(x-\frac{15}{2}=2\) 

x = 19/2

b) \(2012\times x-2010\times x=2014\)

\(x\times\left(2012-2010\right)=2014\)

\(x\times2=2014\)

x = 1007

c) ( x + 1) + (x+2) + (x+3)+...+(x+100) = 5750

\(x\times100+\left(1+2+3+...+100\right)=5750\)

\(x\times100+5050=5750\)

\(x\times100=700\)

x = 7

a, (6x-39) :3= 201

6x-39= 201.3
6x-39= 603
6x= 603+39
6x= 642
x= 642:6
x= 107

\(\left(6x-39\right):3=201\)

\(\Rightarrow6x-39=201.3=603\)

\(\Rightarrow6x=603+39=642\)

\(\Leftrightarrow x=642:6=107\)

Vậy x=107

C X=118                      B x=475          A sai de

1)x=10

2)x=58

3)x=3

4)x=53

t minh nhe

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Leftrightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=-3+1+1+1\)

\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

\(\Leftrightarrow x+2004=0\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\right)\)

<=> x=-2004

a,\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(< =>\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)

\(< =>\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(< =>\left(x+2004\right).\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Do \(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\)

\(=>x+2004=0\)

\(=>x=-2004\)

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Leftrightarrow\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)

\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

\(\Leftrightarrow x=-2004\)

\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+9}{91}+\frac{x+8}{92}+\frac{x+7}{93}\)

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+9}{91}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+7}{93}+1\right)\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{91}+\frac{x+100}{92}+\frac{x+100}{93}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}-\frac{1}{91}-\frac{1}{92}-\frac{1}{93}\right)=0\)

Để ý thấy cụm đằng sau < 0 nên x=-100