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=>( x + x + x +...+x ) + (1+2+3+....+29+30)=1240
=> 31x + ....tính vế phải....=1240
=> ...............Tự mà làm tiếp nhé!
x+(x+1)+...+(x+30)=1240
=> 31x + (1+2+...+30) = 1240
=> 31x + 465 =1240
=> 31x = 775
=> x = 25
ta có
1+2+3+.........+x=5050
=>\(\frac{x.\left(x+1\right)}{2}=5050\)
=>x.(x+1)=5050.2
=>x.(x+1)=10100
=>x.(x+1)=100.101
=>x=100
Ta có : (6 - x)2014 = (6 - x)2015
=> (6 - x)2014 - (6 - x)2015 = 0
<=> (6 - x)2014(1 - 6 - x) = 0
<=> \(\orbr{\begin{cases}\left(6-x\right)^{2014}=0\\1-6-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}6-x=0\\-5-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-5\end{cases}}\)
sory bạn trừng hợp hai mk nhầm :
1 - (6 - x) = 0
=> 1 - 6 + x = 0
=> -5 + x = 0
=> x = 5
Ta có : x + x + 1 + x + 2 + ... + x + 30 = 1240
<=> (x + x + x + ... + x) + (1 + 2 + 3 + .. + 30) = 1240
<=> 31x + 30 = 1240
<=> 31x = 1240 - 30
<=> 31x = 1210
<=> x = 1210/31
a)
x + (x + 1) + (x + 2) + ... + (x + 30) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
(x + x + ... + x) + (1 + 2 + ... + 30) = 1240
(x . [30 - 1 + 1 + 1]) + ([30 + 1] . [30 - 1 + 1] : 2) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
\(a,x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(31x+1+2+3+...+30=1240\)
\(31x+465=1240\)
\(31x=775\)
\(x=25\)
x(x+1)+(x+2)+(x+3)+..+( x+30)
31x+(1+2+3+.....+30)
31x+210=1240
x=31x=1030
=> x = 1030/31
1 + 2 + 3 +... + x = 210
\(\frac{x.\left(x+1\right)}{2}=210\)
x.(x+1)=210.2
x.(x+1)=420
x.(x+1)=20.21
=>x=20
a) x+(x+1)+(x+2)+…+(x+30)=1240
=>x+x+1+x+2+x+3+…+x+30=1240
=>x+x+x+…+x+1+2+3+…+30=1240
Từ 1->30 có: (30-1):1+1=30(số)
=>31.x+(30+1).30:2=1240
=>31.x+31.15=1240
=>31.x+465=1240
=>31.x=1240-465
=>31x=775
=>x=775:31
=>x=25
b) 1+2+3+…+x=210
=>x.(x+1):2=210
=>x.(x+1)=420
=>x.(x+1)=20.21=20.(20+1)
=>x=20
Bài 1:
a,x + ( x + 1) + (x + 2) + (x + 3) +....+ (x + 30) = 1240
x + x +x +.... + x + (1 + 2+ 3+ ....+ 30) = 1240
31x + 465 =1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
b, 1+2+3+...+x=210
\(\frac{x.\left(x+1\right)}{2}=210\)
x(x+1)=210.2
x(x+1)=420
x(x+1)=20.21
=>x=20
\(a.x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(\left(x+x+x+...+x\right)+\left(1+2+...+30\right)=1240\)
\(31x+465=1240\)
\(31x=775\)
\(x=25\)
\(b.\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=6550\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=6550\)
\(100x+5050=6550\)
\(100x=1500\)
\(x=15\)
x+(x+1)+(x+2)+....+(x+30)=1240
<=> (x+x+x+....+x)+(1+2+3+...+30)=1240
<=> 31x+\(\frac{\left(30+1\right)\cdot30}{2}\)=1240
<=> 31x+465=1240
<=> 31x=775
<=> x=25
Vậy x=25