Dạng này rất đơn giản, bạn nhìn các câu hỏi trước của bạn mà làm.

Ta có ; \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\frac{2}{5}\)

\(\Rightarrow x=\frac{2}{5}-\frac{1}{2}=-\frac{1}{10}\)

1)

A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)

A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)

A = \(\frac{1}{1}-\frac{1}{101}\)

A = \(\frac{100}{101}\)

Vậy A = \(\frac{100}{101}\)

B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)

B = \(\frac{5}{2}.\frac{100}{101}\)

B = \(\frac{250}{101}\)

Vậy B = \(\frac{250}{101}\)

2) 

Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )

\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)

\(\Rightarrow d=1\)

Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản

Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

Vậy ...

a) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )

S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)

S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)

S = \(\frac12\) - \(\frac19\)

S = \(\frac{9}{18}-\frac{2}{18}\)

S = \(\frac{7}{18}\)

a) \(\frac{9}{20}\)                                                   c) \(\frac{-55}{4}\)                                                                                                                                 

b) \(\frac{116}{75}\)                                                d) \(\frac{-76}{45}\)

đúng hết đấy nhé mình tính kĩ lắm ko sai đâu

       chúc may mắn

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

a) x=\(\frac{-9}{10}\)

b)x=-6

a

\(5\frac{4}{7}:x+=13\)

\(\frac{39}{7}:x=13\)

\(x=\frac{39}{7}:13\)

\(x=\frac{3}{7}\)

\(\frac{4}{7}x=\frac{9}{8}-0,125\)

\(\frac{4}{7}x=1\)

\(x=1:\frac{4}{7}\)

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

a) \(\left(6\frac{2}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)

=> \(\left(\frac{44}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}=-\frac{11}{7}\)

=> \(\frac{44}{7}x+\frac{3}{7}=-\frac{5}{7}\)

=> \(\frac{44}{7}x=-\frac{8}{7}\)

=> \(\frac{44x}{7}=-\frac{8}{7}\)

=> 44x = -8 => 11x = -2 => \(x=-\frac{2}{11}\)

b) \(3\frac{1}{4}x+\left(-\frac{7}{6}\right)-1\frac{2}{3}=\frac{5}{12}\)

=> \(\frac{13}{4}x-\frac{7}{6}-1\frac{2}{3}=\frac{5}{12}\)

=> \(\frac{13}{4}x-\frac{7}{6}=\frac{25}{12}\)

=> \(\frac{13}{4}x=\frac{13}{4}\)

=> x = 1

c) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

=> \(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

d) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

=> \(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

e) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)

=> \(\left(3x-\frac{7}{9}\right)^3=-1\frac{5}{27}-\left(-\frac{24}{27}\right)=-\frac{32}{27}+\frac{24}{27}=-\frac{8}{27}\)

=> \(\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

=> \(3x-\frac{7}{9}=-\frac{2}{3}\)

=> \(x=\frac{-\frac{2}{3}+\frac{7}{9}}{3}=\frac{1}{27}\)

g) \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{99\cdot100}=\frac{99}{100}\)

=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{99}-\frac{x}{100}=\frac{99}{100}\)

=> \(\frac{x}{1}-\frac{x}{100}=\frac{99}{100}\)

=> \(\frac{100x-x}{100}=\frac{99}{100}\)

=> \(\frac{99x}{100}=\frac{99}{100}\)

=> x = 1

h) \(\frac{x}{3}+\frac{x}{6}+\frac{x}{10}+\frac{x}{15}=3x-1\)

=> \(\frac{2x}{6}+\frac{2x}{12}+\frac{2x}{20}+\frac{2x}{30}=3x-1\)

=> \(\frac{2x}{2\cdot3}+\frac{2x}{3\cdot4}+\frac{2x}{4\cdot5}+\frac{2x}{5\cdot6}=3x-1\)

=> \(2\left(\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}\right)=3x-1\)

=> \(2\left(\frac{x}{2}-\frac{x}{6}\right)=3x-1\)

=> \(2\left(\frac{3x}{6}-\frac{x}{6}\right)=3x-1\)

=> \(2\cdot\frac{2x}{6}=3x-1\)

=> \(\frac{x}{3}=\frac{3x-1}{2}\)

=> 2x = 3(3x - 1)

=> 2x - 9x + 3  = 0

=> -7x = -3

=> x = 3/7

a) 2 - ( \(5\frac{3}{8}\)x  X  - \(\frac{5}{24}\)) = \(\frac{5}{12}\)

             \(5\frac{3}{8}\)x  X  - \(\frac{5}{24}\)= \(\frac{19}{12}\)

             \(5\frac{3}{8}\)x  X             = \(\frac{43}{24}\)

                            X             = \(\frac{1}{3}\)

b)  \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x  X + \(\frac{1}{6}\))  = \(\frac{22}{23}\)

                    \(3\frac{1}{3}\)x  X + \(\frac{1}{6}\)   = \(\frac{23}{18}\)

                   \(3\frac{1}{3}\)x  X                = \(\frac{10}{9}\)

                                  X                 =\(\frac{1}{3}\)

C)  \(\frac{4}{5}\)x    X  - \(\frac{1}{2}\)x    X + \(\frac{3}{4}\)x    X = \(\frac{7}{40}\)

  (   \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\))  x   X                 = \(\frac{7}{40}\)

               \(\frac{21}{20}\)         x   X                      = \(\frac{7}{40}\)

                                       X                       =\(\frac{1}{6}\)