x=2993/225.

\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{41}{10}\div\frac{9}{4}=\frac{41}{10}\times\frac{4}{9}=\frac{82}{45}\Rightarrow\frac{82}{45}=\frac{x}{7,3}\)

Đến đấy thôi

\(\frac{41}{\frac{10}{\frac{9}{4}}}=\frac{x}{7,3}\)

\(\frac{4,1}{2,25}=\frac{x}{7,3}\)

\(x=4,1\cdot7,3\div2,25\)

\(x=\frac{2993}{225}\)

\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\)

\(\Rightarrow x.\frac{9}{4}=\frac{41}{10}.7,3\)

\(\Rightarrow x.\frac{9}{4}=\frac{2993}{100}\)

\(\Rightarrow x=\frac{2993}{100}:\frac{9}{4}\)

\(\Rightarrow x=\frac{2993}{225}\)

Vậy \(x=\frac{2993}{225}.\)

Chúc bạn học tốt!

\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\\ \Rightarrow\frac{9}{4}x=\frac{41}{10}\cdot7,3\left(\text{tính chất của tỉ lệ thức}\right)\\ x=\frac{\frac{41}{10}\cdot7,3}{\frac{9}{4}}=\frac{2993}{225}\)

Vậy \(x=\frac{2993}{225}\)

`1, (10-4-6)+(5/4-9/14-5/7+7/3)

=0+187/84

=187/ 84

29/32.41/36-29/32.32/58-41/36.29/32-41/36.18/41

=-29/32.32/58-41/36.18/41

=-1/2-1/2=-1

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

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

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

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

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

 \(x+\frac{8}{45}=-\frac{37}{45}\)

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

\(x=-1\)

thế phần b, c đâu bạn

\(\frac{15}{41}+\frac{-138}{41}< x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Leftrightarrow\frac{-123}{41}< x< \frac{1.3+1.2+1}{6}\)

\(\Leftrightarrow-3< x< 1\)

\(\Rightarrow x\in\left\{-2;-1;0\right\}\)

\(\frac{x}{5}=\frac{15}{2}-\frac{51}{10}\)

\(\frac{x}{5}=\frac{15.5-51}{10}\)

\(\frac{x}{5}=\frac{24}{10}\)

\(\frac{x}{5}=\frac{12}{5}\)

\(x=12\)

\(\frac{2x}{3}-\frac{1}{9}=\frac{59}{36}+\frac{1}{4}\)

\(\frac{2x}{3}-\frac{1}{9}=\frac{59+9}{36}\)

\(\frac{2x}{3}-\frac{1}{9}=\frac{68}{36}\)

\(\frac{2x}{3}=\frac{68}{36}+\frac{1}{9}\)

\(\frac{2x}{3}=\frac{68}{36}+\frac{4}{36}\)

\(\frac{2x}{3}=2\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\frac{8}{99}\)

\(A=\frac{4}{33}\)

b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)

\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)

c] Ta đặt \(\left(8n+5,6n+4\right)=d\)

\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)

Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản

a, \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)(1)

ĐKXĐ: \(\hept{\begin{cases}x+9\ne0\\x+10\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-9\\x\ne-10\end{cases}}}\)

(1)\(\Leftrightarrow\frac{9.\left(x+9\right)}{90}+\frac{10.\left(x+10\right)}{90}=\frac{9.\left(x+9\right)}{\left(x+9\right)\left(x+10\right)}+\frac{10.\left(x+10\right)}{\left(x+9\right)\left(x+10\right)}\)

\(\Leftrightarrow9.\left(x+9\right)+10.\left(x+10\right)=9.\left(x+9\right)+10.\left(x+10\right)\)

\(\Leftrightarrow9x+81+10x+100=9x+81+10x+100\)

\(\Leftrightarrow9x+10x-9x-10x=81+100-81-100\)

\(\Leftrightarrow0x=0\)

\(\Rightarrow x\in R\)trừ -9 và -10

a,\(=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)

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

=-1+1+-2/11

=0+-2/11

=-2/11

b,\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{40}\right)+\frac{-5}{17}\)

=1+-1+-5/17

=0+-5/17

=-5/17

c,\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+-\frac{7}{9}\right)+\frac{16}{17}\)

=1+-1+16/17

=0+16/17

=16/17

d,\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

a.\(\frac{-5}{9}\)+\(\frac{8}{15}\)+\(\frac{-2}{11}\)+\(\frac{4}{-9}\)+\(\frac{7}{15}\)

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

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

=(-1)+1+\(\frac{-2}{11}\)

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

=\(\frac{-2}{11}\).

ai tk hộ mk đi