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

\(x=\frac{39}{7}:13=\frac{3}{7}\)

\(b,\frac{2}{3}\cdot x+\frac{1}{2}=\frac{1}{10}\)

\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{2}=-\frac{2}{5}\)

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

\(c,x:\frac{8}{11}=\frac{11}{3}\)

\(x=\frac{11}{3}\cdot\frac{8}{11}=\frac{8}{3}\)

\(d,\frac{2}{9}-\frac{7}{8}\cdot x=\frac{1}{3}\)

\(\frac{7}{8}.x=\frac{2}{9}-\frac{1}{3}=-\frac{1}{9}\)

\(x=-\frac{1}{9}:\frac{7}{8}=-\frac{8}{63}\)

cảm ơn Ngyễn Thị Ngọc Ánh nha bạn lm đúng mà mình k nhầm xl bạn nhìu

\(\(a,\frac{39}{7}:x=13\) \(x=\frac{39}{7}:13=\frac{3}{7}\) \(b,\frac{2}{3}\cdot x+\frac{1}{2}=\frac{1}{10}\) \(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{2}=-\frac{2}{5}\) \(x=-\frac{2}{5}:\frac{2}{3}=-\frac{3}{5}\) \(c,x:\frac{8}{11}=\frac{11}{3}\)\)

\(\(x=\frac{11}{3}\cdot\frac{8}{11}=\frac{8}{3}\) \(d,\frac{2}{9}-\frac{7}{8}\cdot x=\frac{1}{3}\) \(\frac{7}{8}.x=\frac{2}{9}-\frac{1}{3}=-\frac{1}{9}\) \(x=-\frac{1}{9}:\frac{7}{8}=-\frac{8}{63}\) \)

\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)

\(\dfrac{1}{6}+x=\dfrac{14}{5}\)

\(=>x=\dfrac{79}{30}\)

\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)

\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)

\(\dfrac{32}{35}:x=\dfrac{11}{6}\)

\(x=\dfrac{192}{385}\)

\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)

\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)

\(x:\dfrac{3}{4}=\dfrac{38}{30}\)

\(=>x=\dfrac{19}{20}\)

\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)

a   1/6+x=14/5

     x=14/5-1/6

     x= 79/30

b    8/5:x=11/6*7/4

     8/5:x=77/24

     x=77/24:8/5

     x=385/192

c  x:3/4=11/3-24/10

  x:3/4=19/5

  x=19/5*3/4

  x=57/20

a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)

=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)

=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)

b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)

=> \(-\frac{7}{8}x-\frac{23}{4}=3\)

=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)

=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)

c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)

=> \(2x-\frac{2}{7}-7=-11\)

=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)

=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)

d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)

=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)

=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)

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

/ là phân số nha

11/13-(5/42-x)=(15/28-11/13)

11/13-(5/42-x)=-37/182

(5/42-x)=11/13+37/182

(5/42-x)=191/182

x=5/42-191/182

x=-254/273

vậy x=-254/273

bạn ấn vào chữ M nằm ngang ở dưới câu hỏi khi bạn trả lời là sẽ ấn được phân số nhé 

còn gửi câu trả lời cũng tương tự 

\(b,x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)

\(< =>x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)

\(< =>x=\frac{1}{3}-\frac{1}{3}+\frac{2}{5}=\frac{2}{5}\)

\(c,\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

\(< =>\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)

\(< =>x=\frac{-29}{70}:\frac{2}{3}=-\frac{7}{140}\)

\(d,\frac{3}{7}-x=\frac{1}{4}-\left(-\frac{3}{5}\right)\)

\(< =>\frac{3}{7}-\frac{1}{4}-\frac{3}{5}=x\)

\(< =>x=-\frac{59}{140}\)

\(e,-\frac{21}{12}x+\frac{1}{3}=-\frac{2}{3}\)

\(< =>-\frac{21}{12}x=-\frac{2}{3}-\frac{1}{3}=-1\)

\(< =>x=\frac{12}{21}\)

\(A=x^2-2x+10\)

\(A=\left(x^2-2x+1\right)+9\)

\(A=\left(x-1\right)^2+9\)

Mà  \(\left(x-1\right)^2\ge0\)

\(\Rightarrow A\ge9\)

Dấu "=" xảy ra khi :

\(x-1=0\Leftrightarrow x=1\)

Vậy Min A = 9 khi x = 1

\(B=x^2-5x-7\)

\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)

\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x-\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow B\ge-\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)

\(C=3x^2+3x-5\)

\(3C=9x^2+9x-15\)

\(3C=\left(9x^2+9x+\frac{9}{4}\right)-\frac{69}{4}\)

\(3C=\left(3x+\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(3x+\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow3C\ge-\frac{69}{4}\)

\(\Leftrightarrow C\ge-\frac{23}{4}\)

Dấu "=" xảy ra khi :

\(3x+\frac{3}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy ...