\(\frac{2,4.1994.2+1,6.3996.3+1,2.4010.4}{3+7+11+15+...+95+99-11.25}\)
p/s: Dấu chấm là dấu nhân nhé.
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15,4 - ( 3 . x + 0,8 ) . 2 = 6
( 3 . x + 0,8 ) . 2 = 15,4 - 6
( 3 . x + 0,8 ) . 2 = 9,4
3 . x + 0,8 = 9,4 : 2
3 . x + 0,8 = 4,7
3 . x = 4,7 - 0,8
3 . x = 3,9
x = 3,9 : 3
x = 1,3
Vậy x = 1,3
15,4 - (3x + 0,8) . 2 = 6
( 3x + 0,8). 2 = 9,4
3x + 0,8 = 4,7
3x = 3,9
=> x = 1,3
Bài 1:
\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)
Bái 2:
\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)
\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)
\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)
Bài 1:
Ta có:
\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)
\(\frac{17}{2}-\left|2x-\frac{5}{2}\right|=-\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{17}{2}-\frac{-7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{51}{6}+\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{29}{3}\)
\(2x-\frac{5}{2}=\frac{29}{3}\)hoặc \(2x-\frac{5}{2}=\frac{-29}{3}\)
Trường hợp 1:
\(2x-\frac{5}{2}=\frac{29}{3}\)
\(2x=\frac{29}{3}+\frac{5}{2}\)
\(2x=\frac{73}{6}\)
\(x=\frac{73}{6}:2\)
\(x=\frac{73}{12}\)
Trường hợp 2:
\(2x-\frac{5}{2}=\frac{-29}{3}\)
\(2x=\frac{-29}{3}+\frac{5}{2}\)
\(2x=\frac{-43}{6}\)
\(x=\frac{-43}{6}:2\)
\(x=\frac{-43}{12}\)
Vậy \(x=\frac{73}{12}\)hoặc \(x=\frac{-43}{12}\)
\(\frac{4,8\left(1994+3996+4010\right)}{\left(\frac{99-3}{4}+1\right)\left(\frac{99+3}{2}\right)-11.25}=\frac{48000}{25.51-11.25}=\frac{48000}{1000}=48\)