Đề thiếu rồi bạn

\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\)

\(\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-1}{12}\)

\(\frac{-13}{9}\le x\le\frac{-1}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{-1}{18}\)

\(\Rightarrow x\in\left(\frac{-26}{18};...;\frac{-18}{18};...\frac{-1}{18}\right)\)

mà x phải thuộc Z

\(\Rightarrow x=\frac{-18}{18}=-1\)

CHÚC BN HỌC TỐT!

Theo đầu bài ta có:
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)\cdot2=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)

\(a,2\frac{x}{7}=\frac{75}{35}\)

\(\frac{75}{35}=2\frac{5}{35}=2\frac{1}{7}\)

\(\Rightarrow x=1\)

\(b,47:4=...\left(+3\right)\)

     \(\Rightarrow\left(47-3\right):4=11\)

      \(\Rightarrow x=11\)

\(c,\frac{112}{5}=\frac{336}{15}=22\left(+6\right)\)

     \(\Rightarrow x=22\frac{x=6}{15}\)

\(d,7,5x=\frac{75x}{100}:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)

     \(\frac{75x}{100}:\left(9-\frac{139}{21}\right)=\frac{63}{25}\)

       \(\frac{75x}{100}:\frac{50}{21}=\frac{63}{25}\)

           \(\Rightarrow\frac{63^3}{25^{\left(1\right)}}\cdot\frac{50^{\left(2\right)}}{21^1}=6=\frac{6}{1}\)

             \(\Rightarrow\frac{6}{1}=\frac{600}{100}\)

Con d sai đề

ra nhiều thế

Giúp mình đi mình kết bạn cho

dễ mà bạn đây là bài cơ bản lớp 6 dấy

 câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
 

a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)

b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)

c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)

d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)

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) \(-\frac{2x}{5}=\frac{-6}{3}\)

\(\Rightarrow-2x.3=-6.5\)

\(\Rightarrow-2x=\frac{-6.5}{3}\)

\(\Rightarrow-2x=-10\)

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

\(\Rightarrow x=5\)

b) \(\frac{x-2}{18}=\frac{-1}{6}\)

\(\Rightarrow\left(x-2\right).6=-1.18\)

\(\Rightarrow x-2=\frac{-1.18}{6}\)

\(\Rightarrow x-2=-3\)

\(\Rightarrow x=-3+2\)

\(\Rightarrow x=-1\)

c) \(\frac{x}{14}=\frac{6}{y}=\frac{-15}{35}\)

\(\Rightarrow\left[\begin{matrix}\frac{x}{14}=\frac{-15}{35}\\\frac{6}{y}=\frac{-15}{35}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x.35=-15.14\\6.35=-15.y\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=\frac{-15.14}{35}\\y=\frac{6.35}{-15}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-6\\y=-14\end{matrix}\right.\)

Vậy \(x=-6;y=-14\)

a) \(\frac{-2x}{5}=-\frac{6}{3}\)

\(\Leftrightarrow-6x=-30\)

\(\Leftrightarrow x=5\)

b) \(\frac{x-2}{18}=-\frac{1}{6}\)

\(\Leftrightarrow-6\left(x-2\right)=-18\)

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=5\)

c) \(\frac{x}{14}=\frac{6}{y}=-\frac{15}{35}\)

\(\Leftrightarrow\frac{x}{14}=\frac{6}{y}=-\frac{3}{7}\)

\(\Leftrightarrow\left\{\begin{matrix}x=-6\\y=-14\end{matrix}\right.\)