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A = \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)
A = 1 - \(\frac{1}{4}\)
A = \(\frac{3}{4}\)
A = 2 + 3/2 + 4/3
A = 2/1 + 3/2 + 4/3
A = 12 / 6 + 9 / 6 + 8 / 6
A = 12/6 + ( 9/6 + 8/6 )
A = 12/6 + 17/6
A = 29/6
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/7-1/8+1/8-1/9
=1/2-1/9
= 7/18
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
=> \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(\left(x+\frac{1}{2\cdot3}\right)+\left(x+\frac{1}{3\cdot4}\right)+....+\left(x+\frac{1}{15\cdot16}\right)=\frac{39}{16}\)
\(\Leftrightarrow\left(x+x+x+....+x\right)+\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{15\cdot16}\right)=\frac{39}{16}\)
\(\Leftrightarrow14x+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{39}{16}\)
\(\Leftrightarrow14x+\left(\frac{1}{2}-\frac{1}{16}\right)=\frac{39}{16}\)
\(\Leftrightarrow14x+\frac{7}{16}=\frac{39}{16}\)
\(\Leftrightarrow14x=2\)
\(\Leftrightarrow x=\frac{1}{7}\)
các bn chỉ cần làm bài 2 thôi nhé!mk biết làm bài 1 rùi.ai làm xong bài 2 trước ngày mai mk tích cho
A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
A=\(\frac{1}{2}-\frac{1}{7}\)
A=\(\frac{5}{14}\)
A = 1/2 -1/3 +1/3-1/4 + 1/4-1/5 +1/5-1/6 + 1/6-1/7 =
1/2-1/7 = 5/14
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{a.\left(a+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\)
\(\Rightarrow a+1=100\)
\(\Rightarrow a=99\)
=>1/2-1/3+1/3-1/4+...+1/a-1/a+1=49/100
=>1/2-1/a+1=49/100
=>1/a+1=49/100+1/2
=>1/a+1=99/100
=>\(\frac{99}{\left(a+1\right).99}=\frac{99}{100}\)
=>(a+1).99=100
=>a+1=100/99
=>a=100/99-1
=>a=1/99