\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)

\(=>6x-5x=20=>x=20\)

áp dụng \(\frac{a}{b}=\frac{c}{d}< =>a.d=b.c\)
 

\(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Rightarrow\left(x+4\right)^2=20.5=100=10^2\)

\(\Rightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)

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 

Có lẽ bạn viết đề sai.

Câu hỏi của Vũ Mai Linh - Toán lớp 7 - Học toán với OnlineMath

mik viết đề đúng mà sai chỗ nào vayh

Mình đang cần gắp 

bạn còn

chỉ có một số \(\frac{-3}{4}\) thôi nha

ĐKXXD : \(x\ne20;8;3;1\)

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{\left(x-1\right)-\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-3\right)-\left(x-8\right)}{\left(x-3\right)\left(x-8\right)}+\frac{\left(x-8\right)-\left(x-20\right)}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}+\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow-\frac{1}{x-1}=-\frac{3}{4}\Leftrightarrow x-1=\frac{4}{3}\Rightarrow x=\frac{7}{3}\)

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

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

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

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

 \(x+\frac{8}{45}=-\frac{37}{45}\)

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

\(x=-1\)

thế phần b, c đâu bạn

a)  \(2\frac{1}{4}x-9\frac{1}{4}=20\)

\(\frac{9}{4}x=20+\frac{37}{4}\)

\(\frac{9}{4}x=\frac{80}{4}+\frac{37}{4}\)

\(\frac{9}{4}x=\frac{117}{4}\)

\(x=\frac{117}{4}:\frac{9}{4}\)

\(x=\frac{117}{4}.\frac{4}{9}\)

\(x=13\)

Vậy x=13

a) \(2\frac{1}{4}x-9\frac{1}{4}=20\)

\(\Rightarrow\frac{9}{4}x-\frac{37}{4}=20\)

\(\Rightarrow\frac{9}{4}x=20+\frac{37}{4}\)

\(\Rightarrow\frac{9}{4}x=\frac{117}{4}\)

\(\Rightarrow x=\frac{117}{4}:\frac{9}{4}\)

\(\Rightarrow x=13\)

Vậy x = 13

b) \(0,25x-\frac{1}{5}x=\frac{13}{20}\)

\(\Rightarrow\left(0,25-\frac{1}{5}\right)x=\frac{13}{20}\)

\(\Rightarrow\frac{1}{20}x=\frac{13}{20}\)

\(\Rightarrow x=\frac{13}{20}:\frac{1}{20}\)

\(\Rightarrow x=13\)

Vậy x = 13

\(b,\)\(0,25x-\frac{1}{5}x=\frac{13}{20}\)

\(\Leftrightarrow x\left(0,25-\frac{1}{5}\right)=\frac{13}{20}\)

\(\Rightarrow x.\frac{1}{20}=\frac{13}{20}\)

\(\Rightarrow x=\frac{13}{20}:\frac{1}{20}\)

\(x=13\)

\(\begin{array}{l}a)x - \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right) = \dfrac{9}{{20}}\\x = \dfrac{9}{{20}} + \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right)\\x = \dfrac{9}{{20}} + \dfrac{{25}}{{20}} - \dfrac{{28}}{{20}}\\x = \dfrac{{6}}{{20}}\\x = \dfrac{{ 3}}{{10}}\end{array}\)

Vậy \(x = \dfrac{{ 3}}{{10}}\)

\(\begin{array}{*{20}{l}}{b)9 - x = \dfrac{8}{7} - \left( { - \dfrac{7}{8}} \right)}\\\begin{array}{l}9 - x = \dfrac{8}{7} + \dfrac{7}{8}\\9 - x = \dfrac{{64}}{{56}} + \dfrac{{49}}{{56}}\\9 - x = \dfrac{{113}}{{56}}\end{array}\\{x = 9 - \dfrac{{113}}{{56}}}\\{x = \dfrac{{504}}{{56}} - \dfrac{{113}}{{56}}}\\{x = \dfrac{{391}}{{56}}}\end{array}\)

Vậy \(x = \dfrac{{391}}{{56}}\)

Vì các phân số đều có dạng  
 
 
 

=>phần này dễ rùi bn tự lm nhé tích trung tỉ ngoại tỉ