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Đề bài thiếu dấu cộng ở giữa ( x + 2 ) và ( x + 3 ) nha bạn.
(x+1)+(x+2)+(x+3)+......+(x+100)=5700
=> x + 1 + x + 2 + ... + x + 100 = 5700
=> (x + x + x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5700
100 số hạng
=> x.100 + 5050 = 5700
=> x.100 = 5700 - 5050
=> x.100 = 650
=> x = 650 : 100
=> x = 6,5
(x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 5700
x + 1 + x + 2 + x + 3 + ... + x + 100 = 5700
100x + 1 + 2 + 3 + ... + 100 = 5700
100x + 5050 = 5700
100x = 5700 - 5050
100x = 650
x = 650 : 100
x = 6,5
( x + 1 ) + ( x + 2 ) + ( x + 3 ) = 100
=> 3x + 6 = 100
=> 3x = 94
=> x = \(\frac{94}{3}\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=100\)
\(\Rightarrow3x+6=100\)
\(\Rightarrow3x=94\)
\(\Rightarrow x=\frac{94}{3}\)
Vậy...
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
\(B=\left(1+\dfrac{1}{100}\right)\times\left(1+\dfrac{1}{99}\right)\times....\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{2}\right)\)
\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)
\(B=\dfrac{101\times100\times....\times4\times3}{100\times99\times....\times3\times2}\)
\(B=\dfrac{101}{2}\)
\(\Rightarrow B=\left(\dfrac{100}{100}+\dfrac{1}{100}\right)\times\left(\dfrac{99}{99}+\dfrac{1}{99}\right)\times...\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)
\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)
\(B=\dfrac{101}{2}\)( triệt tiêu các mẫu, tử giống nhau)
x x 3 + : 0,5=12,8
x x (3 + 2)=12,8
x x 5=12,8
x = 12,8 :5
x = 2,56
\(\left[x+1+x+2+x+3+...+x+100\right]=5950\)
\(\left[\left(x+x+x+...+x\right)+\left(1+2+3+4+...+100\right)\right]=5950\)
\(\left[\left(x+x+x+...+x\right)+\left(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\right)\right]=5950\)
\(\left[100x+5050\right]=5950\)
\(100x=900\Leftrightarrow x=9\)