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\(M=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)
\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(M=\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot10}+\dfrac{5}{10\cdot13}+...+\dfrac{5}{25\cdot28}\)
\(M=\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{25\cdot28}\right)\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}\cdot\dfrac{3}{14}=\dfrac{5}{14}\)
\(A=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{140}\)
\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(3A=5\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-...-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=5\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=5.\dfrac{3}{14}=\dfrac{15}{14}\)
\(\Rightarrow A=\dfrac{15}{14}:3=\dfrac{15}{14}.\dfrac{1}{3}=\dfrac{5}{14}.\)
Vậy \(A=\dfrac{5}{14}.\)
mình ko hiểu chỗ 3A là gì?
Bạn giải thích giúp mình với
\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(\Rightarrow B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(\Rightarrow\frac{3B}{5}=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)
\(\Rightarrow\frac{3B}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\)
\(\Rightarrow\frac{3B}{5}=\frac{1}{4}-\frac{1}{28}\)
\(\Rightarrow\frac{3B}{5}=\frac{3}{14}\)
\(\Rightarrow B=\frac{3}{14}.\frac{5}{3}\)
\(\Rightarrow B=\frac{5}{14}\)
Vậy \(B=\frac{5}{14}\)
Câu 1:
Ta có: \(\dfrac{x-4}{y-3}=\dfrac{4}{3}\)
=> \(3.\left(x-4\right)=4.\left(y-3\right)\)
=>\(3x-12=4y-12\)
=>\(3x=4y\) (1)
Ta có: \(x-y=5\)
=> \(y=y+5\) Thay vào (1) ta có:
\(3.\left(y+5\right)=4.\)y
=>\(3y+15=4y\)
=> \(15=4y-3y\)
=> 15 = y
=> y =15
ta có: x = y +5
=> x = 15 +5
=> x =20
Câu 2:
\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)
\(B=\dfrac{5}{28}+\dfrac{6}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(B=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(B=5,\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(3B=5.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(3B=5.\dfrac{3}{14}\)
\(B=\dfrac{15}{14}:3=\dfrac{5}{14}\)
Câu 3:
38 - (|x+10|+13) = \(\left(-6\right)^{20}:\left(9^9.4^{10}\right)\)
=> \(38-\left(\left|x+10\right|+13\right)=\left(2.3\right)_{ }^{20}:\)\(\left[\left(3^2\right)^9.\left(2^2\right)^4\right]\)
=>\(38-\left(\left|x+10\right|+13\right)=2^{20}.3^{20}:\left(3^{18}.2^{20}\right)\)
=> \(38-\left(\left|x+10\right|+13\right)=\dfrac{3^{20}.2^{20}}{3^{18}.2^{20}}\)
=> \(38-\left(\left|x+10\right|+13\right)=9\)
=> |x +10| + 13 = 38 -9
=> |x+10| +13 = 29
=> |x+10| = 29 -13
=> |x+10| = 16
\(\Rightarrow\left[{}\begin{matrix}x+10=16\\x+10=-16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-26\end{matrix}\right.\)
D= 1/2. (1/25-1/27 +1/27-1/29+...+1/73-1/75)
= 1/2. (1/25 -1/75)
=1/2 . 2/75= 1/75
D = \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+...+\dfrac{1}{73.75}\)
2D = 2( \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+...+\dfrac{1}{73.75}\) )
= \(\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}\)
= \(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)
= \(\dfrac{1}{25}-\dfrac{1}{75}\)
= \(\dfrac{2}{75}\)
b: \(B=2013+\dfrac{2013}{3}+\dfrac{2013}{6}+\dfrac{2013}{10}+\dfrac{2013}{15}\)
\(=2013\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\right)\)
\(=4026\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)
\(=4026\cdot\dfrac{5}{6}=3355\)
\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(=\dfrac{5}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{14}\)
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