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\(\left(x+\dfrac{1}{2\times3}\right)+\left(x+\dfrac{1}{3\times4}\right)+\left(x+\dfrac{1}{4\times5}\right)+\left(x+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)
\(x+\dfrac{1}{2\times3}+x+\dfrac{1}{3\times4}+x+\dfrac{1}{4\times5}+x+\dfrac{1}{5\times6}=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
\(x\times4+\dfrac{4}{12}=\dfrac{25}{3}\)
\(x\times4=\dfrac{25}{3}-\dfrac{4}{12}\)
\(x\times4=\dfrac{25}{3}-\dfrac{1}{3}\)
\(x\times4=\dfrac{24}{3}\)
\(x\times4=8\)
\(x=8\div4\)
\(x=2\)
:))
\(\text{Đặt }A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{49}{100}\)
1/2x3+1/3x4+....+1/99x100
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+....+1/99-1/100
=1-1/100
=99/100
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/19-1/20
=1/2-1/20
=10/20-1/20
=9/20
\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{19\times20}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{5\cdot6}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{5}-\frac{1}{6}\right)=1-\frac{1}{6}=\frac{5}{6}.\)
đặt A=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6
Đặt A = 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
A = 1/1 - 1/6
A = 5/6
Vậy A = 5/6
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)+\left(\dfrac{5}{6}+\dfrac{19}{20}+...+\dfrac{2549}{2550}\right)\)
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{50\cdot51}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{50\cdot51}\right)\)
\(B=\left(1+1+...+1\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\)
\(B=1\cdot49=49\) (vì có (50 - 2) : 1 + 1 = 49 số hạng 1)
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
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