làm tiếp cái trước(ấn nhầm)

\(x=\frac{5}{42}-\frac{15}{28}\) 

\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)

\(x=\frac{20}{168}-\frac{90}{168}\)

\(x=\frac{-70}{168}\)

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

2. 

1.

 \(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)

\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)

\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)

/x+\(\frac{4}{15}\)/ \(-\)\(3,75\)\(=\)\(-2,15\)

/ \(x+\frac{4}{15}\)/ \(=\)\(-2,15+3,75\)

/ \(x+\frac{4}{15}\)/ \(=\)\(\frac{8}{5}\)

TH1: \(x+\frac{4}{15}=\frac{8}{5}\)                                          TH2: \(x+\frac{4}{15}=-\frac{8}{5}\)

\(=>x=\frac{8}{5}-\frac{4}{15}\)                                    \(=>x=-\frac{8}{5}-\frac{4}{15}\)

\(=>x=\frac{24}{15}-\frac{4}{15}\)                                   \(=>x=-\frac{24}{15}-\frac{4}{15}\)

\(=>x=\frac{20}{15}\)                                                 \(=>x=-\frac{28}{15}\)

\(=>x=\frac{4}{3}\)                                                         

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)

B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)

\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)

\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)

\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)

\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)

=> 5x - 20 = 2

=> 5x = 22 

\(\Rightarrow x=\frac{22}{5}=4,4\)

Vậy, x = 4,4

a) 3. (  x - 2/5 ) = 0 

           <=>    x = 0 : 3 + 2/5 

             <=>  x = 2/5 

( Theo cách lớp  6) 

Hoặc cách thông thường nek: 

 3. ( x - 2/5 ) = 0

       x  - 2/5 = 0 : 3

      x  - 2/5 = 0

      x          = 0 + 2/5 

      x         = 2/5

Mấy câu sau làm như vậy nha!

^^ Chúc học tốt!!!!       

Bạn ơi """" 3.x chứ ko phải là 3 đâu

\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{5}{28}-\frac{11}{13}\right)\)

\(\frac{11}{13}-\frac{5}{42}+x=-\frac{5}{28}+\frac{11}{13}\)

\(x=-\frac{5}{28}+\frac{11}{13}-\frac{11}{13}+\frac{5}{42}\)

\(x=\left(\frac{11}{13}-\frac{11}{13}\right)-\left(\frac{5}{28}-\frac{5}{42}\right)\)

\(x=-\left(\frac{15}{84}-\frac{10}{84}\right)\)

\(x=-\frac{5}{84}\)