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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
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cau 1 :
Tu dau bai ta co:
X x 50 + ( 100 + 2 ) x 50 : 2 = 2700
X x 50 + 2550 = 2700
X x 50 = 2700 - 2550
X x 50 = 150
X = 150 : 50
X = 3
Bai 2
1 ) ( 99 + 0 ) x 100 : 2 = 4950
2 ) ( 99 + 1 ) x 99 : 2 = 4950
3 ) day tren co so cac so hang la :
( 99 - 1 ) : 2 + 1 = 50 ( so )
Vay tong la : ( 99 + 1 ) x 50 : 2 = 2500
lik e cho minh nha moi nguoi
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Số lượng số hạng x:
(99-1):2 + 1= 50 (số hạng)
(x+1) + (x+3) + (x+5) +...+ (x+99)= 2450
50x + (1+99) x 25 = 2450 (chia thành 25 cặp)
50x + 2500 = 2450
50x= 2450 - 2500
50x = -50
x= -50:50
x=-1
(x+1)+(x+3)+...+(x+99)=2450
=>99x+\(\dfrac{\left(99-1\right)+1\cdot\left(1+99\right)}{2}\)=2450
=>99x+4950=2450
=>99x=2450-4950=-2500
=>x=-2500:99=\(\dfrac{-2500}{99}\).
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\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\\ \left(x+x+...+x\right)+\left(1+3+5+...+99\right)=0\\ 50x+2500=0\\ x=-50\)
Vậy x = -50
Ta có: \(\left(x+1\right)+\left(x+3\right)+...+\left(x+99\right)=0\)
\(\Leftrightarrow50x=-2500\)
hay x=-50
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`(x+1)+(x+3)+(x+5)+.....+(x+99)=0`
`=>(x+x+.....+x)+(1+3+5+......+99)=0`
`=>50x+(100.50)/2=0` (do từ 1 đến 99 có 50 số nên vì vậy cũng có 50x)
`=>50x+50.50=0`
`=>50(x+50)=0`
`=>x+50=0`
`=>x=-50`
Vậy `x=-50`
(x+1)+(x+3)+(x+5)+...+(x+99)=0
<=> 50x + (1+99) x 25=0
<=>50x+ 2500=0
<=>50x= -2500
<=>x= -50
=>S={-50}
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(x+1)+(x+3)+...+(x+99)=1000
=>50x+1+3+...+99=1000
=>50x+\(\dfrac{\left(99-1\right):2+1\cdot\left(1+99\right)}{2}\)=1000
=>50x+2500=1000
=>50x=1000-2500=-1500
=>x=-1500:50=-30.
\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=1000\)
\(\left(x+x+x+...+x\right)+\left(1+3+5+7+9+...+99\right)\)
\(\left(x.50\right)+2500\)=1000
\(x.50=1000-2500\)
\(x.50=-1500\)
\(x=-1500:50=-30\)
Ta có:
( x + 1 ) + ( x + 3 ) + ( x + 5 ) + .... + ( x + 99 ) = 5 000
=> x + 1 + x + 3 + x + 5 +.....+ x + 99 = 5 000
=> ( x + x + x +....+ x ) + ( 1 + 3 + 5 +.....+ 99 ) = 5 000
=> 50x + 2 500 = 5 000
=> x = ( 5 000 - 2 500 ) : 50
=> x = 50
50x + 2500 = 5000
50x = 5000 - 2500
50x = 2500
x = 2500 : 50
x = 50