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

em đăng bên box toán để mấy bạn kia trả lời nhé =)))))

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

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

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

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

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

Vậy x = -2004

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

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

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

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

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

<=> 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

6 lần số bé là:

201-3=198

Số bé là:

198:6=33

Số lớn là:

201-33=168

Bài 1 :

SL = SB x 5 + 3

Coi số bé là 1 phần thì số lớn là 5 phần và 3 đơn vi

Số bé là : ( 201 - 3 ) : ( 5 + 1 ) = 33

Số lớn là : 201 - 33 = 168

Bài 2 :

x + x . 1/4 : 2/7 + x : 2/9 = 255

x . 1 + x . 1/4 . 7/2 + x . 9/2 = 255

x . ( 1 + 1/4 . 7/2 + 9/2 ) = 255

x . ( 1 + 7/8 + 9/2 ) = 255

x . 51/8 = 255

          x = 255 : 51/8

          x = 40

=168

=40

40793.96757982241

bài này dễ 

bằng 40793.96757982241

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