Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

-x + 20 = - (-15) - (8) + 13 

-x + 20 = 15 - 8 + 13 

-x + 20 = 7 + 13 

- x + 20 = 20

x = 20 - 20 

x = 0

-(-10) + x = -13 + (-9) + (-6) 

10 + x = -13 - 9 - 6 

10 + x = -28 

x = -28 - 10 

x = -38 

8 - (-12) + 10 = -(-14) - x 

8 + 12 + 10 = 14 - x 

30 = 14 - x 

x = 14 - 30 

x = -16

1. -x+20 = -(-15)-8+13

=> -x=15-8+13-20

=> -x=0

=> x=0

2. -(-10)+x=-13+(-9)+(-6)

=> 10+x=-13-9-6

=> x = -13-9-6-10

=> x = -38

3. 8-(-12)+10=-(-14)-x

=> 8+12+10=14-x

=> x = 14-8-12-10

=> x = -16

4. -(+12)+(-x)-(-3)=5-(-7)

=> -12-x+3=5+7

=> -x=5+7+12-3

=> -x=21

=> x=-21

5. 14-x+(-10)=-(-9)+(+15)

=> 14-x-10=9+15

=> -x=9+15-14+10

=> -x=20

=> x=-20

6. 12-(-17)+(-3)=-5+x

=> 12+17-3+5=x

=> x=31

7. x-(-19)-(+32)=14-(+16)

=> x+19-32=14-16

=> x=14-16+32-19

=> x=11

8. x-|-15|-|7|=-(-9)+|-5|

=> x-15-7=9+5

=> x=9+5+7+15

=> x=36

9. 15-x+17=13-(-21)

=> 15-x+17=13+21

=> -x=13+21-15-17

=> -x=2

=> x=-2

10. -|-5|-(-x)+4=3-(-25)

=> -5+x+4=3+25

=> x=3+25-4+5

=> x=29

Bạn xem lại đề nha \(\dfrac{x+2}{15}\)

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

=> \(\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

=> x=-2 do  \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\)\(\ne0\)

\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

\(\Rightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)

\(\Rightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

\(\Rightarrow x+2=0\left(\text{Vì }\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\right)\)

=>x=-2

Vậy x=-2

giải cả ra

x+2/11+x+2/12+x+2/13=x+2/14+x+2/15

=>x+2/11+x+2/12+x+2/13-x+2/14-x+2/15=0

=>(x+2).(1/11+1/12+1/13-1/14-1/15)=0

mà 1/11>1/12>1/13>1/14>1/15

=>1/11+1/12+1/13-1/14-1/15 khác 0

=>x+2=0

=>x=-2

vậy x=-2

NHANH NHANH GIÚP MÌNH VỚI NHÉ MÌNH ĐANG CẦN GẤP 

Bài 7:

a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)

    \(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)

     \(x=\dfrac{21}{55}\)

b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)

    \(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)

      \(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)

      \(x\) = 1

c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)

     \(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)

       \(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)

        24 = \(x\)

EM CẢM ƠN CÔ Ạ

X=-2

 Tick rồi mk ns cách làm cho,hứa

Bài 1:

a) \(\left|3x-5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))

Bài 2:

a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)

b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)

\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)

\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)

Bài 1:

a) \(\left|3x-5\right|=4\)  (1)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)    \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)

\(\Leftrightarrow x=-1\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)           \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

\(\Leftrightarrow x=-2004\)