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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. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)=16:2\)
\(\Leftrightarrow\left(x+1\right)=8\)
\(\Leftrightarrow x=8-1=7\)
\(a)\) \(A=4+2^2+2^3+...+2^{20}\)
\(A=2^2+2^2+2^3+...+2^{20}\)
\(2A=2^3+2^3+2^4+...+2^{21}\)
\(2A-A=\left(2^3+2^3+2^4+...+2^{21}\right)-\left(2^2+2^2+2^3+...+2^{20}\right)\)
\(A=2^3+2^{21}-2^2-2^2\)
\(A=2^3+2^{21}-2.2^2\)
\(A=2^3+2^{21}-2^3\)
\(A=2^{21}\)
Vậy \(A=2^{21}\)
\(b)\) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow\)\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(\Leftrightarrow\)\(100x+\frac{100\left(100+1\right)}{2}=5750\)
\(\Leftrightarrow\)\(100x+5050=5750\)
\(\Leftrightarrow\)\(100x=5750-5050\)
\(\Leftrightarrow\)\(100x=700\)
\(\Leftrightarrow\)\(x=\frac{700}{100}\)
\(\Leftrightarrow\)\(x=7\)
Vậy \(x=7\)
Chúc bạn học tốt ~
Đặt A=1+2+22+23+…+220
=>2.A=2+22+23+24+…+221
=>2.A-A=2+22+23+24+…+221-1-2-22-23-…-220
=>A=221-1
Vậy 1+2+22+23+…+220=221-1
(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
Từ 1 đến 100 có:(100-1):1+1=100(số)
=>100.x+(100+1).100:2=5750
=>100.x+101.50=5750
=>100.x+5050=5750
=>100.x=5750-5050
=>100.x=700
=>x=7
Vậy x=7
A=4+22+23+24+...+220
=22+22+23+24+...+220
=>2A=23+23+24+...+221
=>2A-A=23+23+24+...+221-22-22-23-24-...-220
=>A(2-1)=23+221-22-22
=>A=8+221-4-4
=>A=221
a) x . 100 + (1 + 2 + .... + 100) = 5750
x . 100 + 5050 = 5750
x . 100 = 5750 - 5050
x . 100 = 700
x = 700 : 100
x = 7
b) vô câu hỏi tương tự ấy, lười ghi quá :)))
S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)
\(\left(x+1\right)+\left(x+2\right)+.....+\left(x+100\right)=5750\)
\(\Rightarrow x+1+x+2+.....+x+100=5750\)
\(\Rightarrow100x+1+2+3+....+100=5750\)
\(\Rightarrow100x+\left[\left(\dfrac{100-1}{1}+1\right):2\right]\left(100+1\right)=5750\)
\(\Rightarrow100x+5050=5750\)
\(\Rightarrow100x=700\)
\(\Rightarrow x=7\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=7\)
Vậy ...
Câu 1:
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x102-101x101-51-50\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x\left(102-101\right)-\left(50+51\right)\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101-101\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x0}{2+4+6+8+...+2048}\)
\(A=0\)
Ta có:Số số hạng từ 2 đến 101 là:
(101-2):1+1=100(số hạng)
Do đó từ 2 đến 101 có số cặp là:
100:2=50(cặp)
\(B=\frac{101+100+99+...+3+2+1}{101-100+99-98+3-2+1}\)
\(B=\frac{5151}{51}\)
\(B=101\)
Câu 2:
a)697:\(\frac{15x+364}{x}\)=17
\(\frac{15x+364}{x}\)=697:17
\(\frac{15x+364}{x}\)=41
15x+364=41x
41x-15x=364
26x=364
x=14
Vậy x=14
b)92.4-27=\(\frac{x+350}{x}+315\)
\(\frac{x+350}{x}+315\)=341
\(\frac{x+350}{x}\)=26
x+350=26
x=26-350
x=-324
Vậy x=-324
c, 720 : [ 41 - ( 2x -5)] = 40
[ 41 - ( 2x -5)] =720:40
[ 41 - ( 2x -5)] =18
2x-5=41-18
2x-5=23
2x=28
x=14
Vậy x=14
d, Số số hạng từ 1 đến 100 là:
(100-1):1+1=100(số hạng)
Tổng dãy số là:
(100+1)x100:2=5050
Mà cứ 1 số hạng lại có 1x suy ra có 100x
Ta có:(x+1) + (x+2) +...+ (x+100) = 5750
(x+x+...+x)+(1+2+...+100)=5750
100x+5050=5750
100x=700
x=7
Vậy 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