9/1x2+9/2x3+9/3x4+.................+9/98x99+9/99x100
TÍNH
A= 9/1x2 + 9/2x3 + 9/3x4 +..........+ 9/98x99 + 9/99x100
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+....+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9\cdot\frac{99}{100}=\frac{891}{100}\)
\(A=9\left(\frac{1}{1x2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
=> \(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
=> \(A=9\left(1-\frac{1}{100}\right)=\frac{9.99}{100}=\frac{891}{100}\)
=> A=8,91
Tính giá trị biểu thức:
A=9/1x2 + 9/2x3 + 9/3x4 +...+9/98x99 + 9/99x100
Ta có:\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}=\frac{891}{100}\)
3/1x2-5/2x3+7/3x4-9/4x5+.......-197/98x99
1x2+2x3+3x4+...+98x99+99x100 = ?
Đặt S = 1x2+2x3+3x4+...+98x99+99x100
S x 3 =1x2x3+2x3x3+3x4x3+...+98x99x3+99x100x3
S x 3 =1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+....+98x99x(100-97)+99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4-1x2x3+3x4x5-2x3x4+...+98x99x100-97x98x99+99x100x101-98x99x100
S x 3 = 99x100x101
S x 3 = 999900
S = 333300
S= 1x2+2x3+3x4+4x5+...+98x99+99x100
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
1/1x2 + 1/2x3 + 1/3x4 +......+1/98x99 + 1/99x100
1/1.2 +1/2.3 +1/3.4 +...+1/98.99 +1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100
=1-1/100=100/100-1/100=99/100
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow1-\frac{1}{100}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}=\frac{99}{100}\)
Tính:1/1x2 + 1/2x3 + 1/3x4 + ... + 1/98x99 + 1/99x100
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Cho hai số biết rằng bớt số thứ nhất 28 đơn vị thì được số thứ hai va 1/3 số thứ nhất bằng 3/5 số thứ hai.Tìm hai số đó
Tính A= 1/1x2+1/2x3+1/3x4+........+1/98x99+1/99x100
tính nhanh:5/1x2 + 5/2x3 + 5/3x4 ...+ 5/98x99 + 5/ 99x100
=5(x1/1x2 + 1/2x3 +... +1/99x100)
= 5 x( 1/1 - 1/2 +1/2 -1/3 +... +1/99 -1/100)
= 5 x( 1 /1- 1/100)
= 5 x99/100
= 99/ 20