Toán Lớp 6: Tìm x
(x+1)+(x+2)+….+(x+100)=5050
|1/2+1/3+1/4+…+1/100+x|=-1
1/1.2+1/2.3+1/3.4+…+1/x.(x+1)=1/2
x+2021/2+x+2021/6+x+2021/12=1/2+1/6+1/12
Cần trước 8 rưỡi
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Toán Lớp 6: Tìm x
(x+1)+(x+2)+….+(x+100)=5050
|1/2+1/3+1/4+…+1/100+x|=-1
1/1.2+1/2.3+1/3.4+…+1/x.(x+1)=1/2
x+2021/2+x+2021/6+x+2021/12=1/2+1/6+1/12
Cần trước 8 rưỡi
Comments ( 2 )
\underbrace{(x+x+…+x)}_{100 chữ x}+\underbrace{(1+2+…+100)}_{Đặt là A}=5050
Ta có:A=1+2+3+…+100
A=[(100+1)[(100-1):1+1]]/2
A=[101[99:1+1]]/2
A=[101.[99+1]]/2
A=[101.100]/2
A=[101.50.2]/2
A=101.50
A=5050
=>(x+x+…+x)+A=5050
=>100x+5050=5050
=>100x=5050-5050
=>100x=0
=>x=0:100
=>x=0
Vậy x=0
|1/2+1/3+1/4+…+1/100+x|=-1
Vì |1/2+1/3+1/4+…+1/100+x|\geq 0=>|1/2+1/3+1/4+…+1/100+x|=-1(vô lý)
=> x in \emptyset
Vậy không tìm được giá trị x thỏa mãn yêu cầu bào toán
1-1/2+1/2-1/3+1/3-1/4+…+1/x-1/[x+1]=1/2
1-1/[x+1]=1/2
1/[x+1]=1-1/2
1/[x+1]=2/2-1/2
1/[x+1]=1/2
x+1=2
x=2-1
x=1
Vậy x=1
(x+2021). (1/2+1/6+1/12)=1/2+1/6+1/12
x+2021=(1/2+1/6+1/12):(1/2+1/6+1/12)
x+2021=1
x=2021-1
x=2020
Vậy x=2020