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\(a,345-x=257\)
\(x=88\)
\(b,7.x=115\)
\(x=\frac{115}{7}\)
\(c,\left(x+x+...+x\right)+\left(1+2+...+100\right)=5750\)
\(x\times100+5050=5750\)
\(x\times100=700\)
\(x=7\)
a) A = 4 + 4 + 8 + 16 + ...... + 1048576
2A = 8 + 8 + 16 + ...... + 1048576 + 2.1048576
2A - A = (8 + 8 + 16 + ...... + 1048576 + 2.1048576) - (4 + 4 + 8 + 16 + ...... + 1048576)
A = 2.1048576 + 8 - 4 - 4
A = 2.1048576 = 2097152
b) (x + 1) + (x + 2) + ...... + (x + 100) = 5750
x + 1 + x + 2 + ...... + x + 100 = 5750
100x + (1 + 2 + 3 + ..... + 100) = 5750
Ta có :
1 + 2 + 3 + ..... + 100 = 5050
=> 100x + 5050 = 5750
=> 100x = 200
=> x = 2
Ta có: \(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow100x+5050=5750\)
\(\Leftrightarrow100x=700\)
hay 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
100x = 700
x = 700 : 100
x = 7
x(x+1)+(x+2)+(x+3)+...+(x+100)=5750
=> (x+x+x+x+...+x)+(1+2+3+...+100)=5750
=> 101x+5050=5750
101x = 5750-5050
101x = 700
x = 700 : 101
x = 700/101
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
(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 + 4 + ... + 100) = 5750 (100 số hạng x)
=> 100x + \(\left[\left(100-1\right):1+1\right].\left(\frac{100+1}{2}\right)\) = 5750
=> 100x + \(100.\frac{101}{2}\)= 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
Ta có:
(x + 1) + (x + 2) + ... +(x + 100) = 5750
100x + 1 +2 + 3 + 4 + 5 +... + 100= 5750
100x + \(\frac{1+100}{2}\) = 5750
100x + 1+ 100 = 11500
100x = 11399
x = 113,99
câu 1 có rồi
x-34.15=0
=> x-510=0
=> x=510
18(x-16)=18
=> x-16=18:18
=> x-16=1
=> x=1+16
=> x=17
(x+1)+(x+2)+(x+3)+....+(x+100)=5750