Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt A
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{5}{2}.\left(\frac{12}{25}\right)=\frac{6}{5}\)
Ai thấy đúng thì ủng hộ nha !!!
a, \(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
=\(\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)=\(\frac{5}{2}.\frac{12}{25}\)=\(\frac{6}{5}\)
@@ dùng máy tính mà tính
Anh làm mẫu 1 phần
\(\frac{\frac{2}{2017}+\frac{2}{2018}}{\frac{5}{2017}+\frac{5}{2018}}=\frac{2.\left(\frac{1}{2017}+\frac{1}{2018}\right)}{5.\left(\frac{1}{2017}+\frac{1}{2018}\right)}=\frac{2}{5}\)
Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)
\(B=5\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Rightarrow B=5\cdot\left(1-\frac{1}{103}\right)=5\cdot\frac{102}{103}=\frac{510}{103}\)
\(C=5\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}\right)\)
\(\Rightarrow C=5\cdot\left(1-\frac{1}{101}\right)=5\cdot\frac{100}{101}=\frac{500}{101}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(C=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(C=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(C=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(C=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)
\(1-\frac{1}{x+1}=\frac{4}{5}\)
\(\frac{x}{x+1}=\frac{4}{5}\)
\(\frac{x}{x+1}=\frac{4}{4+1}\)
\(\Rightarrow x=4\)
Vậy x = 4
=))
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{x-1}=1-\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{5}\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
~ Rất vui vì giúp đc bn ~ ^_<
a,8/3x + 26/3= 10/3
8/3x = 10/3- 26/3 = -16/3
=>x = -16/3 : 8/3 = -2
b, (2/3-1/2)x = 5/12
1/6x = 5/12
=>x = 5/2
xong rùi đó
nhớ tk nha
\(c.x\left(\frac{2}{5}-\frac{3}{5}\right)=\frac{2}{35}\)
\(x=\frac{2}{35}:\frac{-1}{5}=-\frac{2}{7}\)
\(d.\left(2x+1\right)^2=49=7^2=\left(-7\right)^2\)
\(TH1:2x+1=7\Rightarrow x=3\)
\(TH2=2x+1=-7\Rightarrow x=-4\)
\(a.x=\frac{-3}{5}-\frac{4}{9}=\frac{-47}{45}\)
\(b.\frac{3}{5}:x=\frac{17}{10}-\frac{2}{5}\)
\(x=\frac{3}{5}:\frac{13}{10}=\frac{6}{13}\)