Câu 1:

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

\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)

Câu 2:

a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)

\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)

=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)

b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)

=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)

=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)

=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)

c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)

=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)

=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)

=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)

Đề bài câu b bài 1 là gì vậy bạn?

|\(\frac32x\) + \(\frac12\)| = |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;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)

b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)

c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)

d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)

lưu ý mk ko chép đầu bài

mình cần gấp lắm đến chiều mai là phải nộp rùi

giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa

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

c)

\(\left|\dfrac{2}{3}x-\dfrac{1}{2}\right|-1=\dfrac{5}{6}\)

\(\Leftrightarrow\left|\dfrac{4x-3}{6}\right|=1+\dfrac{5}{6}=\dfrac{11}{6}\)

\(\Leftrightarrow\left|4x-3\right|=11\Leftrightarrow\left[{}\begin{matrix}4x-3=11=>x=\dfrac{11+3}{4}=\dfrac{7}{2}\\4x-3=-11=>x=\dfrac{-8}{2}=-4\end{matrix}\right.\)

\(\left|x+5\right|=5\)

<=> \(\hept{\begin{cases}x+5=5\\x+5=-5\end{cases}}\)

<=> \(\hept{\begin{cases}x=0\\x=-10\end{cases}}\)

\(\left|x+1\right|+7=10\)

<=> \(\left|x+1\right|=3\)

<=> \(\hept{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)

<=> \(\hept{\begin{cases}x=2\\x=-4\end{cases}}\)

\(\left|x-3\right|-6=5\)

<=> \(\left|x-3\right|=11\)

<=> \(\hept{\begin{cases}x-3=11\\x-3=-11\end{cases}}\)

<=> \(\hept{\begin{cases}x=14\\x=-8\end{cases}}\)

\(\left|x+2\right|-6\left(x-4\right)=20-6x\)

<=> \(\left|x+2\right|-6x+24=20-6x\)

<=> \(\left|x+2\right|=-4\)

<=> \(\hept{\begin{cases}x+2=-4\\x+2=4\end{cases}}\)

<=> \(\hept{\begin{cases}x=-2\\x=2\end{cases}}\)

a) \(|x+5|=5\)

\(\Rightarrow\orbr{\begin{cases}x+5=5\\x+5=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)

Vậy x = 0 hoặc x = -10

b) \(|x+1|+7=10\)

\(\Rightarrow|x+1|=10-7\)

\(\Rightarrow|x+1|=3\)

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

Vậy x = 2 hoặc x = -4

c) \(|x-3|-6=5\)

\(\Rightarrow|x-3|=5+6\)

\(\Rightarrow|x-3|=11\)

\(\Rightarrow\orbr{\begin{cases}x-3=11\\x-3=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=14\\x=-8\end{cases}}\)

Vậy x = 14 hoặc x = -8

d) \(|x+2|-6\left(x-4\right)=20-6x\)

\(\Rightarrow|x+2|-6x+24=20-6x\)

\(\Rightarrow|x+2|=20-6x-24+6x\)

\(\Rightarrow|x+2|=\left(20-24\right)+\left(-6x+6x\right)\)

\(\Rightarrow|x+2|=-4\)

Vì \(|x|\ge0\)mà \(|x+2|=-4\)

\(\Rightarrow\)Không có giá trị x thỏa mãn 

_Chúc bạn học tốt_

Bài 2:

- \(\frac75\) + \(x\) = - \(\frac34\)

\(x\) = - \(\frac34+\frac75\)

\(x=-\frac{15}{20}\) + \(\frac{28}{20}\)

\(x\) = \(\frac{13}{20}\)

Vậy \(x=\frac{13}{20}\)

Bài 3:

- \(\frac74\) + \(\frac14:x\) = \(\frac32\)

\(\frac14:x\) = \(\frac32+\frac74\)

\(\frac14:x\) = \(\frac64+\frac74\)

\(\frac14:x=\frac{13}{4}\)

\(x=\frac14:\frac{13}{4}\)

\(x=\frac14\times\frac{4}{13}\)

\(x=\frac{1}{13}\)

Vậy \(x=\frac{1}{13}\)

a) x = 1

a.X=\(\frac{25}{7}\)

x=\(\frac{4}{5}\):\(\frac{-3}{2}\)

x=\(\frac{-8}{15}\)

Vậy x=\(\frac{-8}{15}\)

b,-4/7x=7/6+(-5)/6

-4/7x=1/3

     x=1/3:(-4)/7

    x=-7/12

Vậy x=-7/12

Bài 1: Tính ( hợp lý nếu có thể )

\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)

\(=-1+1+\dfrac{2}{-5}\)

\(=0+\dfrac{2}{-5}\)

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

\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)

\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)

\(=0+\dfrac{2}{3}\)

\(=\dfrac{2}{3}\)

\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)

\(=-1+1\)

\(=0\)

\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)

\(=\dfrac{1}{6}+\dfrac{-1}{6}\)

\(=0\)

Bài 2: Tìm x,biết:

a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}-\dfrac{2}{3}\)

\(x=\dfrac{2}{15}\)

Vậy \(x=\dfrac{2}{15}\)

b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)

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

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

Vậy \(x=1\)

c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!

\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)

\(x=\dfrac{1}{44}\)

Vậy \(x=\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\)