Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 169 . ( 3x - 9.17 ) + 24 : 3 = 30
169 . ( 3x - 153 ) + 8 = 30
169 . ( 3x - 153 ) = 30 - 8
169 . ( 3x - 153 ) = 22
3x - 153 = 22 : 169
3x - 153 = \(\frac{22}{169}\)
3x = \(\frac{22}{169}+153\)
3x = \(\frac{25879}{169}\)
x = \(\frac{25879}{169}:3\)
x = \(\frac{25879}{507}\)
Vậy \(x=\frac{25879}{507}\)
b) \(\left(\frac{4}{5}:\frac{6}{5}+\frac{1}{5}:\frac{1}{x}\right).30-26=54\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=54+26\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right).30=80\)
\(\left(\frac{2}{3}+\frac{1}{5}.x\right)=80:30\)
\(\frac{2}{3}+\frac{1}{5}.x=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}-\frac{2}{3}\)
\(\frac{1}{5}.x=2\)
\(x=2:\frac{1}{5}\)
\(x=10\)
Vậy \(x=10\)
c) \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{18}\right):12\frac{1}{9}=0\)
\(\frac{1}{2}-\left(\frac{59}{9}+x-\frac{117}{18}\right):\frac{109}{9}=0\)
\(\frac{1}{2}.\left(\frac{59}{9}-\frac{117}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right).\frac{9}{109}=0\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0:\frac{9}{109}\)
\(\frac{1}{2}.\left(\frac{1}{18}+x\right)=0\)
\(\frac{1}{18}+x=0:\frac{1}{2}\)
\(\frac{1}{18}+x=0\)
\(x=0-\frac{1}{18}\)
\(x=\frac{-1}{18}\)
Vậy \(x=\frac{-1}{18}\)
d) 720 : [ 41 - ( 2x - 5 ) ] = 210
41 - 2x + 5 = 720 : 210
41 + 5 - 2x = \(\frac{24}{7}\)
46 - 2x = \(\frac{24}{7}\)
2x = \(46-\frac{24}{7}\)
2x = \(\frac{298}{7}\)
x = \(\frac{298}{7}:2\)
x = \(\frac{149}{7}\)
Vậy \(x=\frac{149}{7}\)
\(A=-\frac{550}{9}\)\(\Rightarrow\)12.5% của A là \(\frac{-275}{36}\)
\(B=\frac{2}{5}\Rightarrow\)5% của B là \(\frac{1}{8}\)
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
a) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right):2\)
\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=5.\left(1-\frac{1}{101}\right):2=5.\frac{100}{101}:2=\frac{500}{101}.\frac{1}{2}\)\(=\frac{250}{101}\)
b) \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)\(.\frac{1}{3}\)
\(=(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{33}).\frac{1}{3}=\frac{10}{33}.\frac{1}{3}=\frac{10}{99}\)
Ta có : \(A=\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{468}\)
\(\Leftrightarrow\frac{1}{3}A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{468}+\frac{5}{1458}\)
\(\Leftrightarrow A-\frac{1}{3}A=\frac{5}{2}-\frac{5}{1458}\)
\(\Leftrightarrow\frac{2}{3}A=\frac{5}{2}-\frac{5}{1458}\)
\(\Leftrightarrow\frac{2}{3}A=\frac{1820}{729}\)
\(\Leftrightarrow A=\frac{1820}{729}.\frac{3}{2}=\frac{910}{243}\)
dùng máy tính là xong