tìm x thuộc n , biết
(x+1) + ( x + 2 ) + ( x + 3 ) + ... + ( x+ 100 ) = 5750
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Ta có: \(\left(x+1\right)+\left(x+2\right)+...+\left(x+1000\right)=5750\)
\(\Leftrightarrow1000x+500500=5750\)
\(\Leftrightarrow x=-\dfrac{1979}{4}\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(100x+\left(1+2+3+4+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=700\div100\)
\(x=7\)
Vậy ...
( X + 1 ) + ( X + 2 ) + ( X + 3 ) + ... + ( X + 100 ) = 5750
100 X + ( 1 + 2 + 3 + ... + 100 ) = 5750
100 X + 5050 = 5750
100 X = 5750 - 5050
100 X = 700
X = 700 : 100
X = 7
Vậy x = 7
a) \(???\)
b) \(123x+877x=2000\)
\(1000x=2000\)
\(x=2000:1000\)
\(x=2\)
c) \(2x.\left(x-10\right)=0\)
=> \(x-10=0\)
\(x=10\)
d)\(6.\left(x+2\right)-\left(4x+10\right)=100\)
\(6.x+12-4x+10=100\)
\(2x+2=100\)
\(2x=98\)
\(x=98:2\)
\(x=49\)
e) \(x.\left(x+1\right)=2+4+6+8+...+2500\)
\(x.\left(x+1\right)=1563750\)
mà ta thấy : \(1250.1251=1563750\)
=> \(x=1250\)
g)\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(x.100+5050=5750\)
\(x.100=5750-5050\)
\(x.100=700\)
\(x=7\)
x+x+x+...x+ 1+2+3+...100 = 5750
100.x + 101.50 = 5750
100.x = 5757-5050
100.x = 700
x = 700: 100
x = 7
BÀi 2
( x+ 1 )+ ( x +2 ) + ... + ( x + 100) = 5750
x + 1 +x + 2 + .. x+ 100 = 5750
(x+ x+ .. +x ) + ( 1+ 2 + ... +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 + 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
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 + 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
100x +(1+2+3+...+100)=5750
100x + (100+1)x100 :2 = 5750
100x + 10100:2=5750
100x+ 5050 =5750
100x=5750=5050=700
x=700 :100 = 7