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=1-1/2+1/2-1/3+1/3-1/4+1*4-1/5+1/5+1/6=1-1/6=5/6
ok xong
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=1-\frac{1}{2}+\)+.... \(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 1\) ( điều phải chứng minh )
Vậy \(A< 1\)
Chúc bạn học tốt ~
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\left(\text{đ}pcm\right)\)
vậy:\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}< 1\)
k mk bạn nha:)
= \(\left(1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{38}-\frac{1}{38}+\frac{1}{39}\right)\)
= 1 + \(1+\frac{1}{39}=\frac{40}{39}\)
chỗ " 1 + " phía trước là bỏ
ngay chỗ dấu bằng thứ hai
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Rightarrow k=2\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
A = 9(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\))
A = 9(\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\))
A = 9(1 - \(\frac{1}{100}\))
A = 9.\(\frac{99}{100}\)=\(\frac{891}{100}\)=8,91
Vì \(\frac{1}{1.2}=\frac{1}{2}\)
Mà \(\frac{1}{1}-\frac{1}{2}=\frac{2}{2}-\frac{1}{2}=\frac{1}{2}\)
Nên trong bài toán: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
Mấy cái kia cũng vậy nên bạn yên tâm nha!!!!
A : 9 = 1/1.2 + 1/2.3 + 1/3.4 + ..... + 1/98.99 + 1/99.100
A : 9 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/98 - 1/99 + 1/99 - 1/100
A : 9 = 1 - 1/100
A : 9 = 100/100 - 1/100
A : 9 = 99/100
A = 9 . 99/100
A = 891/100 = 8,91 = 8 91/100
Ta có : \(A=2+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(\Rightarrow A=2+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=2+\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=2+\frac{99}{100}=\frac{299}{100}\)
Ta có : A=\(2+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow A=2+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=2+\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=2+\frac{99}{100}\)
\(\Rightarrow A=\frac{299}{100}\)
Can you k for me,Natsu drangeel!