a, 10+15+20+....+295+x.300+x=67

10+15+20+...+295+x(300+1)=67

10+15+20+...+295+x.301=67

8845+x.301=67

67-8845=x.301

-8878=x.301

x=-29/149/301

b, 

\(\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{6}{7}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{1}{x+1}=\frac{6}{7}-\frac{59}{77}\)

\(\frac{1}{x+1}=\frac{1}{11}\)

suy ra x+1=11

suy ra x=10

 \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)

\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)

\(x+A=\frac{47}{42}\)

     ta thấy :

              \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

              \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

                  \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

               \(A=\frac{1}{1}-\frac{1}{6}\)

             \(A=\frac{5}{6}\)

       vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)       

  hay  \(x+\frac{5}{6}=\frac{47}{42}\)

     \(x=\frac{47}{42}-\frac{5}{6}\)

 \(x=\frac{2}{7}\)

\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)

\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)

\(x=\frac{47}{42}-\frac{5}{6}\)

\(x=\frac{2}{7}.\)

\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)

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

\(\Leftrightarrow x=2\)

\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)

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

\(\Rightarrow x=2\)

Vậy \(x=2\)

\(\left(x+50\%\right):\frac{7}{8}=\frac{5}{7}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)=\frac{5}{7}.\frac{7}{8}\)

\(\Rightarrow x+\frac{1}{2}=\frac{5}{8}\)

\(\Rightarrow x=\frac{5}{8}-\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{8}\)
Vậy...

Mình làm tiếp bài của bạn " I have a crazy idea "

b) \(\frac{25-x}{3}=\frac{15}{2}\)

Áp dụng tỉ lệ thức:

\(\left(25-x\right).2=15.3\)

\(\Rightarrow25-x=\frac{15.3}{2}=\frac{45}{2}\Leftrightarrow x=25-\frac{45}{2}=\frac{5}{2}\)

c) \(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}=1\)

\(\Rightarrow x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=1\)

\(\Rightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)

\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)=1\)

\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{7}\right)=1\Leftrightarrow x-\frac{6}{7}=1\Leftrightarrow x=1+\frac{6}{7}=\frac{13}{7}\)

a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath

b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến

c)$x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}$

\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}-\frac{1}{3}\)

\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{-2}{21}\)

\(\Rightarrow\frac{x}{3}=\frac{-2}{21}\div\frac{1}{42}\)

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

\(\Rightarrow\frac{x}{3}=\frac{-12}{3}\)

\(\Rightarrow x=-12\)

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

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