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a) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+..+100\right)=5750\Rightarrow x.100+\left(100+1\right)\cdot100:2=5750\)\
\(\Rightarrow x.100+5050=5750\Rightarrow x.100=700\Rightarrow x=7\)
b) \(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\Rightarrow\left(x+1\right)^2=4^2\)
\(\Leftrightarrow x+1=4\Rightarrow x=3\)
1.\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(\Leftrightarrow100x+5050=5750\)
\(\Leftrightarrow100x=5750-5050=700\)
\(\Leftrightarrow x=700:100=7\)
2. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)=16:2\)
\(\Leftrightarrow\left(x+1\right)=8\)
\(\Leftrightarrow x=8-1=7\)
(x+1)+(x+2)+(x+3)+...+(x+100)=5750
x + 1 + x + 2 + x + 3 + .... + x + 100 = 5750
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750
100x + 5050 = 5750
100x = 5750 - 5050 = 700
x = 700 : 100 = 7
(x+1)+(x+2)+(x+3)+...+(x+100)=5750
x + 1 + x + 2 + x + 3 + .... + x + 100 = 5750
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750
100x + 5050 = 5750
100x = 5750 - 5050 = 700
x = 700 : 100 = 7
(x +1) + (x + 2) + (x + 3) + (x + 100) = 7450
100x + (1 + 2 + 3 + ... + 100) = 7450
100x + 5050 = 7450
100x = 7450 - 5050
100x = 2400
x = 2400 : 100
x = 24
Ta có: (x + 1) + (x + 2) +...+ (x + 100) = 5750
=> (x + x + x +...... +x) + (1 + 2 + 3 + 4 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
Ta có: ( x+1) + (x+2) + (x+3) +...+ ( x+100) = 5750
<=> ( x + x + x + ...... + x ) + (1 + 2 + 3 + ..... + 100) = 5750
<=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
[ x + 1 ] + [ x + 2 ] + [ x + 3 ] + ... + [ x + 100 ] = 5750
[ x + x + x + .... + x ] + [ 1 + 2 + 3 + ... + 100 ] = 5750
Đặt A = 1 + 2 + 3 + ... + 100 và B = x + x + x + .... + x
Dãy A có số số hạng là :
( 100 - 1 ) : 1 + 1 = 100 ( số hạng )
=> A = ( 1 + 100 ) x 100 : 2 = 5050
=> B = 100x
Ta có :
[ x + x + x + .... + x ] + [ 1 + 2 + 3 + ... + 100 ] = 5750
100 x + 5050 = 5750
100 x = 700
x = 7
(x+1)+(x+2)+...+(x+100) = 5750
(x+x+x+.....+x) + ( 1+2+..+100) = 5750
100x + 5050 = 5750
100x = 5750 - 5050
100x = 700
x = 700 : 100
x = 7
100 . x + ( 1 + 2 + 3 + ....+ 100 )
100 . x + 100 . 101 : 2
=> 100x + 101 + 50 = 5750
100x = 5750 - 5050 = 700
=> x = 700 : 100=7
=> x = 7
a) Dãy trên có số số hạng là :
( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x
=> 1 + 2 + 3 + 4 + ... + x = 45
=> ( x + 1 ) . x : 2 = 45
=> ( x + 1 ) . x = 90
=> ( x + 1 ) . x = 10 . 9
=> x = 9
Vậy x = 9
a) Dãy trên có số số hạng là :
( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x
=> 1 + 2 + 3 + 4 + ... + x = 45
=> ( x + 1 ) . x : 2 = 45
=> ( x + 1 ) . x = 90
=> ( x + 1 ) . x = 10 . 9
=> x = 9
Vậy x = 9
x+(x+1)+(x+2)+...(x+100)=8080
(x+x+x+..+x)+(0+1+2+...+100)=8080
101x+5050=8080
101x=8080-5050
101x=3030
x=3030:101
x=30
x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) +.....+ ( x + 100 ) = 8080
=( x + x + x + x +....+ x ) + ( 1 + 2 + 3 + 4 + .... + 100) = 8080
=(x + x + x + ... + x ) + 5050 = 8080
101x = 8080 - 5050
101x = 3030
x = 3030 : 101
x = 30