Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(D=\frac{3}{3x4}+\frac{3}{4x5}+.....+\frac{3}{99x100}.\)
\(D=3x\left(\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}+\frac{1}{99x100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(D=1-\frac{3}{100}\)
\(D=\frac{97}{100}\)
\(D=\frac{3}{3x4}+\frac{3}{4x5}+.........+\frac{3}{98x99}+\frac{3}{99x100}\)
\(D=3x\left(\frac{1}{3x4}+\frac{1}{4x5}+...........+\frac{1}{98x99}+\frac{1}{99x100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(D=\frac{3x97}{100}\)
\(D=\frac{291}{100}\)
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
a) \(\left(x-25\right):15=20\)
\(\Rightarrow x-25=20\times15\)
\(\Rightarrow x-25=300\)
\(\Rightarrow x=300+25\)
\(\Rightarrow x=325\)
Vậy x = 325
b) \(3\times x-25=80\)
\(\Rightarrow3\times x=80+25\)
\(\Rightarrow3\times x=105\)
\(\Rightarrow x=105:3\)
\(\Rightarrow x=35\)
Vậy x = 35
c) \(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{2}-\frac{1}{100}\)
\(S=\frac{49}{100}\)
Vậy \(S=\frac{49}{100}\)
_Chúc bạn học tốt_
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)
\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
\(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{4}-\frac{1}{100}\)
=\(\frac{6}{25}\)
=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
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 = 99x100x101 A = 333300
D= 3/3x4+3/4x5+...+3/99x100
D=3x(1/3x4+1/4x5+....+1/99x100)
D=3x(1/3-1/4+1/4-1/5+...+1/99-1/100)
D=3x(1/3-1/100)
D=3x(100/300-3/300)
D=3x97/300=97/100
Nhớ tk cho mình nha
\(D=\frac{3}{3\times4}+\frac{3}{4\times5}+...+\frac{3}{98\times99}+\frac{3}{99\times100}\)
\(=3\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)
\(=3\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3\times\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(=3\times\frac{97}{300}\)
\(=\frac{97}{100}\)