x ( x + 1) + ( x + 2) + ( x + 3) + ... + ( x + 100) = 25250
Tìm a
a + 2 x a + 3 x a + ...... + 99 x a +100 x a = 25250
<=> a + 2a + 3a + .... + 99a + 100a = 25250
<=> a.(1+ 2 + 3 + .... + 99 + 100) = 25250
<=> a. 5050 = 25250
<=> a = 25250 : 5050
<=> a = 5
a + 2 x a + 3 x a + ...... + 99 x a + 100 x a = 25250
( 1 + 2 + 3 + ... + 99 + 100 ) x a = 25250
5050 x a = 25250
a = 25250 : 5050
a = 5
a + 2 x a + 3 x a + ..... + 99 x a + 100 x a = 25250
= 1 x a + 2 x a + 3 x a + .... 99 x a + 100 x a = 25250
= a x ( 1 + 2 + 3 + ... + 99 + 100 ) = 25250
= a x 5050 = 25250
= a = 25250 : 5050 = 5
Vậy a = 5.
Tìm a
a + 2 x a + 3 x a + ...... + 99 x a +100 x a = 25250
Giúp mk với
a(1+2+3+...+100)=25250
a((1+100).100)/2=25250
a.5050=25250
a=5
a + 2 * a + 3 * a + ...+ 99 * a + 100 * a = 25250
a x 1 + 2 x a + 3 x a + ... + 99 x a + 100 x a = 25250
Số số hạng từ 1 đến 100 là:
( 100 - 1 ) : 1 + 1 = 100 (số)
a x ( 1 + 2 + 3 +... + 99 + 100 ) = 25250
Tổng các số từ 1 đến 100 là:
( 100 + 1 ) x 100 : 2 = 5050
a x 5050 = 25250
a = 25250 : 5050
a = 5
Vậy số cần tìm là 5.
Kb với mình nha
1 x 2 + 2 x 3 + 3 x 4 + ... + 100 x 101 =
1 x 2 x 3 + 2 x 3 x 4 + ... + 100 x 101 x 102
Có cả lời giải nhé
a: S=1(1+1)+2(1+2)+...+100(1+100)
=1+2+...+100+1^2+2^2+...+100^2
\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)
=343450
b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)
=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)
=>4*A=100*101*102*103
=>A=25*101*102*103
4 x 2 : 2 x 3 : 3 x 4 : 4 x........x 100 : 100 + 6 x 1 : 1 x 2 : 2 x 3 : 3 x 4 : 4 x........x 1000 : 1000 x4
giúp nhanh
Tính
a) (x-1/2)+(x-1/4)+(x-1/8)+...+(x-1/512)
Tìm x
a) (x-1/1×2)+(x-1/2×3)+...+(x-1/100×101)
b) (x-1)+(x-2)+(x-3)+...+(x-101)=5050
c) x+1/2+1/3+1/4+...+1/100=3/2+4/3+5/4++...+101/100
1/Tìm x,biết:
a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555
b)1+2+3+4+...+x=820
c)3(x+1)=9.27
d)x+2x+3x+...+99x+100x=15150
e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550
f)3x+3x+1+3x+2=351
a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555
=> 101x +5050 = 5555
=> 101x = 505
=> x = 505 : 101 = 5
Vậy, x = 5
b)1+2+3+4+...+x=820
=> ( x+1) x :2 = 820
=> (x+1)x = 1640
Mà 1640 = 40 . 41
=> x = 40 ( vì {x+1} - x = 1)
Vậy, x = 40
c) 3x+1 = 9.27=243
=> 3x+1 = 35
=>x + 1 = 5
=> x = 4
Vậy, x=4
d) x+2x+3x+...+99x+100x=15150
=> [( 100 + 1) x 100 :2 ] x = 15150
=> 5050x = 15150
=> x = 15150:5050 = 3
Vậy, x =3
e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050= 200500
=> x = 200500 : 100 = 2005
Vậy, x = 2005
f)3x+3x+1+3x+2=351
=> 3x + 3x . 3 + 3x x 9 = 351
=> 3x ( 1+3+9) = 351
=> 3x . 13 = 351
=> 3x = 351 :13=27 mà 27 = 33
=> x=3
Vậy, x=3
a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)
\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)
\(\Rightarrow101\cdot x+5050=5555\)
\(\Rightarrow101\cdot x=5555-5050\)
\(\Rightarrow101\cdot x=505\)
\(\Rightarrow x=505:101\)
\(\Rightarrow x=5\)
b) \(1+2+3+4+...+x=820\)
\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)
\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)
\(\Rightarrow\left(x+1\right)\cdot x:2=820\)
\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)
\(\Rightarrow x\cdot\left(x+1\right)=1640\)
Ta thấy: \(40\cdot41=1640\)
Vậy: \(x=40\)
Bài 6 Tìm x
1) ( x+1 ) + ( x+2 ) + ( x+3 ) + ... + ( x+100 ) = 5750
2) ( 2x-1 ) + ( 4x-2 ) + ... + ( 200x - 100 ) = 5050
3) ( x+2 ) + ( x+4 ) + ( x+6 ) + ... + ( x + 100) = 2650
100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100
A=100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100
A/100=1/1 x 2 + 1/2 x 3 + 1/3 x 4 +...+1/99 x 100
A/100=2-1/1x2 + 3-2/2x3 + ... + 100-99/99x100
A/100=1-1/2 + 1/2-1/3+...+1/99-1/100
A/100=1-1/100
A/100=99/100
A=99/100x100=99
Vậy A=99.
Ta có:
\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)
\(\Rightarrow100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{100}\right)\Leftrightarrow100.\frac{99}{100}=99\)
\(\text{Ta có :}\)
\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)
\(=100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(100.\left(\frac{1}{1}-\frac{1}{100}\right)=100.\frac{99}{100}=99\)
A.(x+1)+(x+2)+(x+3)+......(x+100)=5750
B. x+(1+2+3+4+5+..........+100)=2000
C.(x-1)+(x-2)-(x-3)+(x-4)+........(x-100)=50
Giúp mình với ạ mai mình phải nộp bài rồi
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. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)