a, \(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ mà, bn nhân chia trước, cộng trừ sau, sau đó phá ngọăc rồi tính
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{13}{15}-\frac{7}{10}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{26}{30}-\frac{21}{30}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{1}{6}\)
\(\frac{13}{21}+x=\frac{1}{6}:\frac{7}{12}\)
\(\frac{13}{21}+x=\frac{1.12}{6.7}\)
\(\frac{13}{21}+x=\frac{2}{7}\)
\(x=\frac{2}{7}-\frac{13}{21}\)
\(x=\frac{6}{21}-\frac{13}{21}\)
\(x=\frac{-1}{3}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{13}{15}-\frac{7}{10}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{1}{6}\)
\(\frac{13}{21}+x=\frac{1}{6}:\frac{7}{12}\)
\(\frac{13}{21}+x=\frac{2}{7}\)
\(x=\frac{2}{7}-\frac{13}{21}\)
\(x=-\frac{1}{3}\)