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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 + [x + 1] + [x + 2] + ... + [x + 30] = 1240
=> 31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25
b) 1 + 2 + 3 + ... + x = 210
=< x . (x + 1) = 201 . 2
=> x . (x + 1) = 420
Vì 420 = 20 . 21 nên x = 20
a)(x+x+...+x)+(1+2+...+30)=1240
31x+465=1240
31x=1240-465
31x=775
x=775/31
x=25
b)1+2+3+...+x=210
Đặt A=1+2+3+...+x
A có: (x-1)+1=x(số hạng)
A=(x+1)*x/2=210
(x+1)*x=210*2
(x+1)*x=420
=>x=20
a) \(x+\left(x+1\right)+......+\left(x+30\right)=1240\\ \Rightarrow31x+465=1240\\ \Rightarrow31x=775\\ \Rightarrow x=25\)
b) \(1+2+3+.........+x=210\\ \Rightarrow\frac{\left(x+1\right)x}{2}=210\\ \Rightarrow\left(x+1\right)x=420\\ \Rightarrow x=20\)
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
x = 775 : 31
x = 25
b) 1 + 2 + 3 + ... + x = 210
(x + 1) . (x - 1 + 1) : 2 = 210
(x + 1) . (x - 1 + 1) = 210 . 2
(x + 1) . (x - 1 + 1) = 420
(x + 1) . x = 420
Mà 20 . 21 = 420 => x = 20
a, x + ( x+ 1) + ( x + 2 )+...+( x + 30 )= 1240
1240=31x + (1+2+...+30)
1240=31x +{[(30-1)+1]:2} .(30+1)
1240=31x +15.31
1240=31x+465
31x=1240-465
31x=775
x=775 : 31
x=25
vay x=25
k mình nha
1. Tìm x
a) 1+2+3+...+x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x = 20
b) \(32.3^x=9.3^{10}+5.27^3\)
=>\(32.3^x=9.3^{10}+5.3^9\)(\(27^3=\left(3^3\right)^3=3^9\))
=>\(32.3^x=9.3.3^9+5.3^9\)
=>\(32.3^x=3^9\left(9.3+5\right)\)
=>\(32.3^x=3^9.32\)
=>x = 9
2.
Ta có 2A = 3A - A
=> 2A = \(3\left(1+3+3^2+3^3+....+3^{10}\right)\)\(-\)\(1-3-3^2-3^3-....-3^{10}\)
=> 2A = \(3+3^2+3^3+.....+3^{11}-\)\(1-3-3^2-3^3-...-3^{10}\)
=> 2A = \(3^{11}-1\)
=> 2A+1 = \(3^{11}-1+1\)=\(3^{11}\)
=> n = 11
Ta có : a)1 + 2 + 3 + ... + x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 420
=> x(x + 1) = 20.21
=> x = 20
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
câu a )
x = 25
câu b)
x=20
tk mk nha
a) x + (x + 1) + (x + 2) + ..... + (x + 30) = 1240
31x + (1 + 2 + 3 + ..... + 30) = 1240
31x + 465 = 1240
31x = 775
=> x = 25
b) 1 + 2 + 3 + ..... + x = 210
Áp dụng công thức tính tổng dãy số , ta có :
\(\frac{\left[\left(x-1\right):1+1\right].\left(x+1\right)}{2}=\frac{x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 420
<=> x(x + 1) = 20.21
=> x = 20