Tính giá trị của biểu thức sau bằng cách hợp lí nhất
\(b=\dfrac{4}{8.13}+\dfrac{4}{13.18}+\dfrac{4}{18.23}+...+\dfrac{4}{253.258}\)
hộ mình nhé
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\(\left\{{}\begin{matrix}x⋮12\\x⋮21\end{matrix}\right.\) =>\(x\in BC\left(12;21\right)\)
Ta có: 12= \(2^2\cdot3\)
21 = 3*7
=> BCNN(12;21)= 22 *3*7= 84
BC(12;21)=B(84)={0;84;168;252;336;420;504;....}
Vì 18<x<500 =>x\(\in\left\{84;168;252;336;420\right\}\)
\(5.5.125^2=5.5.\left(5^3\right)^2=5.5.5^{3.2}=5^1.5^1.5^6\\ =5^{1+1+6}=5^8\)
A = { 0;1;2;3;4;5}
A= { x\(\in\) \(ℕ\) | x \(\le\) 5}
\(a=\dfrac{5}{10.12}+\dfrac{5}{12.14}+\dfrac{5}{14.16}+...+\dfrac{5}{998.1000}\\ \dfrac{2}{5}a=\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{998.1000}\\ \dfrac{2}{5}a=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{998}-\dfrac{1}{1000}\\ \dfrac{2}{5}a=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\\ a=\dfrac{99}{400}\)
\(a=\dfrac{5}{10.12}+\dfrac{5}{12.14}+...+\dfrac{5}{998.1000}\)
\(a=5\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+.....+\dfrac{1}{998}+\dfrac{1}{1000}\right)\)
\(A=5\left(\dfrac{1}{10}-\dfrac{1}{1000}\right)=\dfrac{99}{200}\)
17.85 + 15.17-120
= 17. ( 85 + 15) - 120
= 17 . 100 - 120
= 1700 - 120
= 1580
21-(25-x+21)=-(25-7)
21-(25-x+21)=-18
25-x+21 =21+18
25-x+21 =39
25-x =39-21
25-x =18
x =25-18
x =7
vậy x=7
\(\dfrac{5b}{4}=\dfrac{5}{8.13}+\dfrac{5}{13.18}+\dfrac{5}{18.23}+...+\dfrac{5}{253.258}\\ =\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\\ =\dfrac{1}{8}-\dfrac{1}{258}=\dfrac{125}{1032}\\ =>b=\dfrac{25}{258}\)
\(B=\dfrac{4}{8\cdot13}+\dfrac{4}{13\cdot18}+\dfrac{4}{18\cdot23}+...+\dfrac{1}{253\cdot258}\)
\(B=4\left(\dfrac{1}{8\cdot13}+\dfrac{1}{13\cdot18}+\dfrac{1}{18\cdot23}+...+\dfrac{1}{253\cdot258}\right)\)
\(B=\dfrac{4}{5}\left(\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}+...+\dfrac{5}{253\cdot258}\right)\)
\(B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
\(B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)\)
\(B=\dfrac{4}{5}\cdot\dfrac{125}{1032}=\dfrac{25}{258}\)