a/ \(\frac{x-1}{9}=\frac{8}{3}\) 

\(\Leftrightarrow3\left(x-1\right)=72\)

\(\Leftrightarrow x-1=24\)

\(\Leftrightarrow x=25\)

Vậy ..

b/ \(\frac{-x}{4}=\frac{-9}{x}\)

\(\Leftrightarrow x^2=36\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

Vậy ..

c/ \(\frac{x}{4}=\frac{18}{x+1}\)

\(\Leftrightarrow x\left(x+1\right)=72\)

\(\Leftrightarrow x\left(x+1\right)=8.9\)

\(\Leftrightarrow x=8\)

Vậy ..

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Xét vế phải

\(\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\frac{9}{1}\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+9\)

\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\left(1+1+...+1\right)\)

\(=\left(1+\frac{1}{9}\right)+\left(1+\frac{2}{8}\right)+\left(1+\frac{3}{7}\right)+\left(1+\frac{4}{6}\right)+\left(1+\frac{5}{5}\right)+\left(1+\frac{6}{4}\right)+\left(1+\frac{7}{3}\right)+\left(1+\frac{8}{2}\right)+1\)\(=\frac{10}{9}+\frac{10}{8}+\frac{10}{7}+\frac{10}{6}+\frac{10}{5}+\frac{10}{4}+\frac{10}{3}+\frac{10}{2}+\frac{10}{10}\)

\(=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)

Thay vào bài ,ta được:

\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\times x=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)\(\Rightarrow x=10\)

Vậy x=10

chúc bạn học tốt

\(\frac{4}{5}-\left|3-x\right|=\frac{1}{4}\)

\(\Rightarrow\left|3-x\right|=\frac{4}{5}-\frac{1}{4}\)

\(\Rightarrow\left|3-x\right|=\frac{11}{20}\)

\(\Rightarrow\hept{\begin{cases}3-x=\frac{11}{20}\\3-x=-\frac{11}{20}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{49}{20}\\x=\frac{71}{20}\end{cases}}\)

\(\frac{8}{9}:x^2+\frac{1}{9}=1\)

\(\Rightarrow\frac{8}{9}:x^2+\frac{1}{9}=\frac{9}{9}\)

\(\Rightarrow\frac{8}{9}:x^2=\frac{9}{9}-\frac{1}{9}\)

\(\Rightarrow\frac{8}{9}:x^2=\frac{8}{9}\)

\(\Rightarrow x^2=\frac{8}{9}:\frac{8}{9}\)

\(\Rightarrow x^2=1\)

\(\Rightarrow x^2=1^2\)

\(\Rightarrow x=1\)

Thêm:

\(\frac{4}{5}+\frac{3}{4}.x\frac{5}{2}\)

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

a)

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

\(x.\frac{7}{9}=\frac{19}{6}\)

\(x=\frac{19}{6}:\frac{7}{9}\)

\(x=\frac{57}{14}\)

b) \(\frac{5}{7}+x:\frac{9}{4}=\frac{4}{3}\)

\(x:\frac{9}{4}=\frac{4}{3}-\frac{5}{7}\)

\(x:\frac{9}{4}=\frac{13}{21}\)

\(x=\frac{13}{21}.\frac{9}{4}\)

\(x=\frac{39}{28}\)

A,27/14

bài 1: tìm x:

a) x·79 =23 +212

=>x./7/9=19/6

=>x=19/6:7/9

=>x=57/14 

 b) 57 +x:94 =43 

=>x:9/4=4/3-5/7

=>x:9/4=13/21

=>x=13/21x9/4

=>x=39/28

\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)

\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)

\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau

Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

tự làm để nc trí óc dễ

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)

  \(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)

  \(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)

 \(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)

 \(x+\frac{8}{45}=-\frac{37}{45}\)

\(x=-\frac{37}{45}-\frac{8}{45}\)

\(x=-1\)

thế phần b, c đâu bạn

tập trung sẽ bt làm

\(\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-x-\frac{1}{4}\)

\(\frac{4}{15}+\frac{1}{6}-\frac{4}{9}-\frac{2}{3}+\frac{1}{4}>-x\)

\(-\frac{77}{180}>-x\)

\(x>\frac{77}{108}\)