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\(\frac{1}{2\times3}+\frac{1}{3\times4}+............+\frac{1}{a\times\left(a+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..............+\frac{1}{a}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{a+1}=\frac{1}{2}-\frac{49}{100}\)
\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\)
\(\Rightarrow a+1=100\)
\(\Rightarrow a=99\)
Đáp số là a = 99 nha còn cách làm thì Nguyễn Hung Phat đã làm rồi nha
T ik nha bạn =))
Chúc bạn học tốt nhé !!!
\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)
\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)
\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)
\(\Rightarrow100n-100=98n+98\)
\(\Rightarrow2n=198\)
=> n = 99
Vậy n = 99
\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)
\(\frac{1}{2}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2}\)-\(\frac{49}{100}\)
\(\frac{1}{n+1}\)=\(\frac{1}{100}\)
=> n+1=100
n=100-1
n=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 = 1/2-49/100
1/a+1=1/100
a+1=100
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 = 1/2-49/100
1/a+1=1/100
a+1=100
a=99
\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{a\times\left(a+1\right)}=\frac{49}{100}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{\left(a+1\right)}=\frac{49}{100}\)
\(\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)
\(\frac{50}{100}-\frac{1}{a+1}=\frac{49}{100}\)
\(\frac{1}{a+1}=\frac{50}{100}-\frac{49}{100}\)
\(\frac{1}{a+1}=\frac{1}{100}\)
\(\Rightarrow a+1=100\)
\(\Rightarrow a=100-1\)
\(a=99\)
Số a chính là: 99
Chúc bạn may mắn......mình chính là Đào Minh Tiến!
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
=\(1-\frac{1}{9}\)
=\(\frac{8}{9}\)
OK XONG NHỚ CHO MIK NHA
\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)
=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
=1-\(\frac{1}{9}\)
=\(\frac{8}{9}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{9x10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\left(1:a+2a+...+10a\right)=\frac{49}{100}\)
\(\Rightarrow1-10a=\frac{49}{100}\)
\(\Rightarrow10a=1-\frac{49}{100}\)
10a=0,51
a=\(\frac{0,51}{10}=0,051\)
