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\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)
\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
Tìm x :
2,8 : x - 1,5 = 2,3 : 5,75
=> x - 1,5 = 5,75
x = 5,75 + 1,5
x = 7,25
Tính nhanh :
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{13\times14}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{13}-\frac{1}{14}\)
\(=1-\frac{1}{14}+0+...+0\)
\(=\frac{13}{14}\)
1. 2,8 : x - 1,5 = 2,3 : 5,75
2,8 : x - 1,5 = 0,4
2,8 : x = 0,4 + 1,5
2,8 : x = 1,9
x = 2,8 : 1,9
x = 1,473
2. \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{13.14}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)
= \(1-\frac{1}{14}\)
= \(\frac{13}{14}\)
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
Bài này khi sáng mình mới học 100% là đúng luôn.
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ........... + 1/99x100.
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..........1/98-1/99+1/99-1/100.
=1/1-1/100=99/100.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
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_
a) (x-25):15=20
x-25=20×15
x-25=300
x=300+25
x=3325
b) 3×x-25=80
3×x=80+25
3×x=105
x=105:3
x=35
S=1/2×3 + 1/3×4 + 1/4×5 + ... + 1/98×99 + 1/99×100
S=1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
S=1/2-1/100
S=50/100 - 1/100
S=49/100
Tìm x:
a) (x - 25) : 15 = 20
=> x - 25 = 20 x 15
=> x - 25 = 300
=> x = 300 + 25
=> x = 325
Vậy x = 325
b) 3x - 25 = 80
=> 3x = 80 + 25
=> 3x = 105
=> x = 105 : 3
=> x = 35
Vậy x = 35
Tính nhanh tổng sau:
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(\Rightarrow S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow S=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow S=\frac{49}{100}\)
Tìm x
a) ( x - 25 ) : 15 = 20
( x - 25 ) = 20 x 15
( x - 25 ) = 300
x = 300 + 25
x = 325
b) 3 x x - 25 = 80
3 x x = 80 + 25
3 x x = 105
x = 105 : 3
x = 35
Tính nhanh tổng sau
\(S=\frac{1}{2\text{ x }3}+\frac{1}{3\text{ x }4}+\frac{1}{4\text{ x }5}+...+\frac{1}{98\text{ x }99}+\frac{1}{99\text{ x }100}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{2}-\frac{1}{100}\)
\(S=\frac{49}{100}\)
a) (x - 25) : 15 = 20
x - 25 = 20 x 15
x - 25 = 300
x = 300 + 25
x = 325
b) 3 x x - 25 = 80
3 x x = 80 + 25
3 x x = 105
x = 105 : 3
x = 35
S= 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + ... + 1/98 x 99 + 1/99 x 100
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/98 - 1/99 + 1/99 -1/100
= 1/2 - 1/100
= 49/100
a) ( x - 25) : 15 = 20
x - 25 = 300
x = 325
b) 3 * x - 25 = 80
3* x = 105
x = 35
\(S=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}+\frac{1}{99x100}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{2}-\frac{1}{100}\)
\(S=\frac{49}{100}\)