Tìm X
X-1/(1x3)-1/(3x5)-...-1/(97x99)=1/2
1/1x3+1/3x5+......+1/97x99 = ?
\(\frac{1}{1x3}+\frac{1}{3x5}+....+\frac{1}{97x99}\)=S
\(2S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+...+\frac{99-97}{97x99}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
\(S=\frac{2S}{2}=\frac{49}{99}\)
1/1x3+1/2x4+1/3x5+...+1/97x99+1/98x100-49/99
\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+..+\dfrac{1}{97.99}+\dfrac{1}{98.100}-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{99.100}\right)\right]-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{98}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[1-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[\dfrac{98}{99}+\dfrac{49}{100}\right]-\dfrac{49}{99}=\dfrac{14651}{19800}-\dfrac{49}{99}=\dfrac{49}{200}\)
\(\dfrac{1}{1x3}+\dfrac{1}{2x4}+...+\dfrac{1}{98x100}+\dfrac{1}{97x99}-\dfrac{49}{99}=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{100}-\dfrac{49}{99}=1-\dfrac{1}{100}-\dfrac{49}{99}\)
=\(\dfrac{4901}{9900}\)
1/1x3 + 1/3x5 + 1/5x7 + .... + 1/97x99
Đặt S là biểu thức trên
\(\Rightarrow S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{97.99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(\frac{99}{99}-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow S=\frac{49}{99}\)
Vậy biểu thức trên có giá trị là \(\frac{49}{99}\)
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{97\times99}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+....+\frac{1}{97\times99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\frac{98}{99}\)
\(=\frac{49}{99}\)
\(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{97x99}\)
\(=\frac{1}{2}x(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99})\)
\(=\frac{1}{2}x\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}x\frac{98}{99}\)
\(=\frac{49}{99}\)
a)1/1x3+1/3x5+1/5x7+...+1/Xx(x+3)=99/200
b)1/1x3+1/3x5+1/5x7+...+1/Xx(x+2)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)
* Bạn làm theo công thức và vẫn dụng câu b nhé.
Chứng minh : A = 1/1x3 + 1/2x4 + 1/3x5 + 1/4x6 + ... + 1/97x99 + 1/98x100 < 3/4
Ta có :
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
Cho S=1/1x3+1/2x4+1/3x5+..............+1/97x99+1/98x100
So sánh S với 1
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(S< 1-\frac{1}{99}< 1\)
=> S < 1
Tính tổng:
A=1/1+1/2+1/3+1/4+1/5+1/6
B=1x3+3x5+5x7+7x9+...+95x97+97x99
tính tổng
A=1/1x3+1/3x5+....+1/97x99
cần gấp nha
thanks
\(A=\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{97.99}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
so sánh A và B biết
A=\(\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{99x100}\)
B=\(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{97x99}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)