a)\(\frac{2}{3}:\frac{4}{9}+\frac{1}{3}:\frac{4}{9}\)

=\(\frac{4}{9}:\left(\frac{2}{3}+\frac{1}{3}\right)\)

=\(\frac{4}{9}:1=\frac{4}{9}\)

b

a 9/4

b 49/40

c 36/35

d 144/441

a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)

\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)

\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)

\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)

\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)

\(\frac{92}{17}:x=\frac{1}{17}\)

\(x=\frac{92}{17}:\frac{1}{17}\)

\(x=92\)

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=1-\frac{18}{19}\)

\(\frac{1}{x+3}=\frac{1}{19}\)

\(\Rightarrow x+3=19\)

\(\Rightarrow x=19-3\)

\(\Rightarrow x=16\)

(7x6=5+2+6x7) =

Ta có : \(5\frac{8}{17}\div X+\left(-\frac{1}{17}\right)\div X+3\frac{1}{17}\div17\frac{1}{3}=\frac{4}{17}\)

Nên:    \(\left(5\frac{8}{17}+\left(-\frac{1}{17}\right)\right)\div X+\frac{52}{17}\div\frac{52}{3}=\frac{4}{17}\)

           \(5\frac{7}{17}\div X+\frac{52}{17}\times\frac{3}{52}=\frac{4}{17}\)

           \(\frac{92}{17}\div X+\frac{3}{17}=\frac{4}{17}\)

           \(\frac{92}{17}\div X=\frac{4}{17}-\frac{3}{17}\)

           \(\frac{92}{17}\div X=\frac{1}{17}\)

           \(X=\frac{92}{17}\div\frac{1}{17}\)

            \(X=92\)

Vậy \(X=92\)

\(5\frac{8}{17}:x+\left(-\frac{1}{17}\right):x=\frac{52}{51}\)

\(\left(5\frac{8}{17}+-\frac{1}{17}\right):x=\frac{52}{51}\)

\(\frac{92}{17}:x=\frac{52}{51}\)

\(X=\frac{92}{17}:\frac{52}{51}=\frac{69}{13}\)

\(\frac{93}{17}:x+\left(-\frac{1}{17}\right):x+\frac{52}{17}:\frac{52}{3}=\frac{4}{7}\)

\(\frac{93}{17x}+\frac{-1}{17x}+\frac{3}{17}=\frac{4}{7}\)

\(\frac{93+\left(-1\right)}{17x}=\frac{47}{119}\)

\(\frac{92}{17x}=\frac{47}{119}\Rightarrow92\times119=47\left(17x\right)\)

=>10948=799x

=>x=\(\frac{644}{47}\)

\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)

\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)

\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)

\(=-\frac{7}{9}.5\)

\(=-7\)

a)Bn Kaito Kid làm rùi!

B)Không viết lại đề

\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)

c)Không viết lại đề

\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)

\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)

D)

Dòng trên chữ D) là phần c)

Dưới phần c) là phần d) nha!(Lỗi đánh máy)

a) \(15+2\left|x\right|=-3\\ \\ < =>2\left|x\right|=15-\left(-3\right)\\ < =>2\left|x\right|=18\\ =>\left|x\right|=\frac{18}{2}=9\\ =>x=9hoặcx=-9\)

b) \(\left|x-2\right|=7\\ < =>x-2=7hoặcx-2=-7\\ =>x=9hoặcx=-5\)

c) \(100-4.x^2=224\\ < =>4.x^2=100-224=-124\\ < =>x^2=-\frac{124}{4}=-31\\ Mà:x^2\ge0\\ =>xkhôngcógiátrịnàothoảmãn\)

d)\(2x-\frac{9}{240}=\frac{39}{80}\\ < =>2x-\frac{3}{80}=\frac{39}{80}\\ =>2x=\frac{39}{80}+\frac{3}{80}=\frac{21}{40}\\ =>x=\frac{\frac{21}{40}}{2}=\frac{21}{80}\)

\(15+2\left|x\right|=-3\)

\(\Rightarrow2\left|x\right|=15+3\)

\(\Rightarrow2\left|x\right|=18\)

\(\Rightarrow\left|x\right|=\frac{18}{2}\)

\(\Rightarrow\left|x\right|=9\)

\(\Rightarrow\left[\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

Vậy, x = 9 hoặc x = -9.

\(\left|x-2\right|=7\)

\(\Rightarrow\left|x\right|=9\)

\(\Rightarrow\left[\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

a) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)

\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)

\(\Leftrightarrow\frac{3}{5}.x=1\)

\(\Leftrightarrow x=1:\frac{3}{5}\)

\(\Leftrightarrow x=\frac{5}{3}\)

Vậy : \(x=\frac{5}{3}\)

b) \(\frac{4}{7}+\frac{5}{7}:x=1\)

\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)

\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)

\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)

\(\Leftrightarrow x=\frac{5}{3}\)

Vậy : \(x=\frac{5}{3}\)

c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)

\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)

\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)

\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)

\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)

Vậy : \(x=-\frac{19}{12}\)

d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)

\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)

\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)

\(\Leftrightarrow x=\frac{273}{40}\)

Vậy : \(x=\frac{273}{40}\)

\(\)