\(M=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)

\(M=\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot10}+\dfrac{5}{10\cdot13}+...+\dfrac{5}{25\cdot28}\)

\(M=\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{25\cdot28}\right)\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}\cdot\dfrac{3}{14}=\dfrac{5}{14}\)

bằng 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 \(A=\dfrac{5}{14}.\)

\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(\Rightarrow B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(\Rightarrow\frac{3B}{5}=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)

\(\Rightarrow\frac{3B}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\)

\(\Rightarrow\frac{3B}{5}=\frac{1}{4}-\frac{1}{28}\)

\(\Rightarrow\frac{3B}{5}=\frac{3}{14}\)

\(\Rightarrow B=\frac{3}{14}.\frac{5}{3}\)

\(\Rightarrow B=\frac{5}{14}\)

Vậy \(B=\frac{5}{14}\)

\(D=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+....+\dfrac{5}{700}\)

\(=5\left(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+...+\dfrac{1}{700}\right)\)

\(=5\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{25.28}\right)\)

\(=5.\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(=\dfrac{5}{3}.\dfrac{3}{14}\)

\(=\dfrac{5}{14}\)

\(D=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}.\dfrac{6}{28}=\dfrac{5}{14}\)

\(E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)=\dfrac{2.23}{50}=\dfrac{23}{25}.\)

\(\dfrac{D}{E}=\dfrac{5}{24}.\dfrac{25}{23}=\dfrac{125}{552}.\)

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

\(\Rightarrow\left[{}\begin{matrix}x+10=16\\x+10=-16\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-26\end{matrix}\right.\)

Câu 2:

\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+....+\dfrac{10}{1400}\)
\(\Rightarrow B=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+.....+\dfrac{20}{2800}\)
\(\Rightarrow B=20\left(\dfrac{1}{112}+\dfrac{1}{280}+\dfrac{1}{520}+...+\dfrac{1}{2800}\right)\)
\(\Rightarrow B=20\left(\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20.6}{6.56}\)
\(\Rightarrow B=\dfrac{20}{56}\)
\(\Rightarrow B=\dfrac{5}{14}\)

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( \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+...+\dfrac{1}{73.75}\) )

= \(\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}\)

= \(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)

= \(\dfrac{1}{25}-\dfrac{1}{75}\)

= \(\dfrac{2}{75}\)

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