Bài 1:

a) x + \(\frac{1}{9}\) - \(\frac{3}{5}\) = \(\frac{3}{6}\)                                                                                     b) \(\frac{3}{4}\) - x + \(\frac{6}{11}\) = \(\frac{5}{6}\)

    x + \(\frac{1}{9}\)       = \(\frac{3}{6}\) + \(\frac{3}{5}\)                                                                                   \(\frac{3}{4}\) - x         = \(\frac{5}{6}\) - \(\frac{6}{11}\)

    x + \(\frac{1}{9}\)       = \(\frac{11}{10}\)                                                                                         \(\frac{3}{4}\) - x         = \(\frac{19}{66}\)

    x              = \(\frac{11}{10}\) - \(\frac{1}{9}\)                                                                                     x             = \(\frac{19}{66}\) + \(\frac{3}{4}\)

    x              = \(\frac{89}{90}\)                                                                                            x             = \(\frac{137}{132}\)

Bài 2

a) x : 13/16 = 5/8                                                               b)x - 14/28 = 6/9 + 8/25

   x               = 5/8 * 13/16                                                    x - 14/28 = 74/75

   x               = 65/128                                                          x              = 74/75 + 14/28

                                                                                            x               = 223/150

Bài 3

a)62/7 * x = 29/9 : 3/56                                      b)1/5 : x = 1/5 + 1/7                         

   62/7 * x = 1624/27                                              1/5 : x = 12/35

   x           = 1624/27 : 62/7                                            x  = 12/35 * 1/5

  x            = 5684/637                                                    x  = 12/175

Bài 1:

a ) \(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)

\(x+\frac{1}{9}=\frac{3}{6}+\frac{3}{5}\)

\(x+\frac{1}{9}=\frac{11}{10}\)

\(x=\frac{11}{10}-\frac{1}{9}=\frac{89}{90}\)

Vậy \(x=\frac{89}{90}\)

b) \(\frac{3}{4}-x+\frac{6}{11}=\frac{5}{6}\)

\(\frac{3}{4}-x=\frac{5}{6}-\frac{6}{11}\)

\(\frac{3}{4}-x=\frac{19}{66}\)

\(x=\frac{3}{4}-\frac{19}{66}=\frac{61}{132}\)

Vậy \(x=\frac{61}{132}\)

Bài 2 :

a) \(x:\frac{13}{16}=\frac{5}{-8}\)

\(x=\frac{5}{-8}.\frac{13}{16}=-\frac{65}{128}\)

Vậy \(x=-\frac{65}{128}\)

b) \(x.\frac{-14}{28}=\frac{6}{-9}-\frac{2}{15}\)

\(x.\frac{-14}{28}=-\frac{4}{5}\)

\(x=-\frac{4}{5}:\frac{-14}{28}=\frac{8}{5}\)

Vậy \(x=\frac{8}{5}\)

2/ 1 - 3 + 5 - 7 + ... + (x-2) - x = -600

=> ( - 2 ) + ( - 2 ) + .... + ( - 2 ) = - 600 ( số số hạng các số - 2 là A ) 

=> ( - 2 ) . A =  - 600 => A = 300 

Mặt Khác số số hạng ở dãy trên = ( x - 1 ) : 2 + 1 ( 1 ) 

Số số cặp = Số số hạng : 2 = 300 => Số số hạng = 600 

Thay vào ( 1 ) Ta có :

( x - 1 ) : 2 + 1 = 600 => ( x - 1 ) : 2 = 599 => x = 599 . 2 + 1 = 1198+ 1 = 1199 

Các câu 1,3,4 làm tương tự

2+4+6+..........+x=600

4.x=16

-37 + 14 +26 + 37 

=-37 + 37 + 14 + 36 

=0 + 50

=50

  -24 + 6 + 10 + 24 

=(-24 + 24) +(6+10)

= 0+16

= 16

nhiều quá :((

\(a,2\left(x-5\right)-3\left(x+7\right)=14\)

\(2x-10-3x-21=14\)

\(-x-31=14\)

\(-x=45\)

\(x=45\)

\(b,5\left(x-6\right)-2\left(x+3\right)=12\)

\(5x-30-2x-6=12\)

\(3x-36==12\)

\(3x=48\)

\(x=16\)

\(c,3\left(x-4\right)-\left(8-x\right)=12\)

\(3x-12-8+x=0\)

\(4x-20=0\)

\(4x=20\)

\(x=5\)

Cố nốt nha bn ! 

cảm ơn, bn nha:)))

mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???

a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)

\(x=\dfrac{13}{52}+\dfrac{8}{52}\)

⇒ \(x=\dfrac{21}{52}\)

b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)

\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)

\(\dfrac{x}{3}=\dfrac{11}{21}\)

⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)

⇒ \(x=\dfrac{11}{7}\)

c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)

\(\dfrac{-17}{7}< x< -1\)

⇒ \(-17< x< -7\)

⇒ \(x\in\left\{-16;-15,....;-6\right\}\)

d) \(\dfrac{1}{6}+\dfrac{2}{5}\)

\(=\dfrac{5}{30}+\dfrac{12}{30}\)

\(=\dfrac{17}{30}\)

e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)

\(=\dfrac{12}{20}+\dfrac{-35}{20}\)

\(=\dfrac{-23}{20}\)

f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)

\(=\dfrac{4}{13}+\dfrac{-2}{5}\)

\(=\dfrac{20}{65}+\dfrac{-26}{65}\)

\(=\dfrac{-6}{65}\)

g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)

\(=\dfrac{-6}{58}+\dfrac{16}{58}\)

\(=\dfrac{10}{58}\)

h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)

\(=\dfrac{1}{5}+\dfrac{-4}{5}\)

\(=\dfrac{-3}{5}\)

j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)

\(=\dfrac{-2}{9}+\dfrac{5}{9}\)

\(=\dfrac{3}{9}\)

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

a: ta có: \(7-\left(2x-\frac13\right)^2=3\)

=>\(\left(2x-\frac13\right)^2=7-3=4\)

=>\(\left[\begin{array}{l}2x-\frac13=2\\ 2x-\frac13=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=2+\frac13=\frac73\\ 2x=-2+\frac13=-\frac53\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac73:2=\frac76\\ x=-\frac53:2=-\frac56\end{array}\right.\)

b: \(\left(2x+\frac13\right)^2-\frac38=\frac18\)

=>\(\left(2x+\frac13\right)^2=\frac38+\frac18=\frac48=\frac12\)

=>\(2x+\frac13=\sqrt{\frac12}=\frac{\sqrt2}{2}\)

=>\(2x=\frac{\sqrt2}{2}-\frac13=\frac{3\sqrt2-2}{6}\)

=>\(x=\frac{3\sqrt2-2}{12}\)

c: \(12:\left\lbrack29-\left(x-\frac23\right)^2\right\rbrack=3\)

=>\(29-\left(x-\frac23\right)^2=12:3=4\)

=>\(\left(x-\frac23\right)^2=29-4=25\)

=>\(\left[\begin{array}{l}x-\frac23=5\\ x-\frac23=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5+\frac23=\frac{17}{3}\\ x=-5+\frac23=-\frac{13}{3}\end{array}\right.\)

d: \(\left(3x-\frac12\right)^3+\frac83=\frac{29}{9}-\frac{14}{27}\)

=>\(\left(3x-\frac12\right)^3+\frac83=\frac{87}{27}-\frac{14}{27}=\frac{73}{27}\)

=>\(\left(3x-\frac12\right)^3=\frac{73}{27}-\frac83=\frac{73}{27}-\frac{72}{27}=\frac{1}{27}=\left(\frac13\right)^3\)

=>\(3x-\frac12=\frac13\)

=>\(3x=\frac12+\frac13=\frac56\)

=>\(x=\frac{5}{18}\)

e: \(2\left(2x-\frac13\right)^2+\frac43=\frac56+\frac{13}{18}\)

=>\(2\left(2x-\frac13\right)^2=\frac{15}{18}+\frac{13}{18}-\frac43=\frac{28}{18}-\frac{24}{18}=\frac{4}{18}=\frac29\)

=>\(\left(2x-\frac13\right)^2=\frac19\)

=>\(\left[\begin{array}{l}2x-\frac13=\frac13\\ 2x-\frac13=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac23\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=0\end{array}\right.\)