Question

[Nguyễn Lê Hoàng]

Tìm x:

a) x + 3.x + 5.x + .... + 2009.x = 2010. 1005

b) x + (x+1) + (x+2) + .... + (x+30) = 620

Answer 1

[I don't need you]

a) x + 3.x + 5.x + ... + 2009. x = 2010.1005

\(x\times\left(1+3+5+...+2009\right)=2010\times1005\)

\(x\times\left[\left(1+2009\right)\times1005:2\right]=2010\times1005\)

\(x\times2010\times1005\times\frac{1}{2}=2010\times1005\)

\(\Rightarrow x\times\frac{1}{2}=2010\times1005:\left(2010\times1005\right)\)

\(x\times\frac{1}{2}=1\)

x = 2

b) x + (x+1) + (x+2) +...+ (x+30) = 620

x. 31 + ( 1+2+...+30) = 620

x.31 + [ ( 30+1).30:2) = 620

x.31 + 465 = 620

x.31 = 620 - 465

x.31 = 155

x  = 155 : 31

x  = 5

Answer 2

[Dương Lam Hàng]

a) x+3.x+5.x+.....+2009.x = 2010.1005

=> x.(1+3+5+....+2009) = 2010.1005

=> x.1010025 = 2020050

=> x               = 2

Vậy x = 2

b) x+(x+1)+(x+2)+....+(x+30) = 620

=> (x+x+x+...+x)+(1+2+3+...+30) = 620

=> 31x + 465 = 620

=> 31x           = 155

=> x               = 5