Tim x thuoc Z
a}[x+1]+[x+2]+[x+3]+...............+[x+99]=0
b}[x-3]+[x-2]+[x-1]+.................+10+11=11
1 )Tim x, y thuoc Z
x + y = x.y
2) Tim x thuoc Z
(x + 1)+(x+3)+(x+5)+...+(x+99)=0(x-3)+(x-2)+(x-1)+...+10+11=11-12(x-5)+7(3-x)=530(x+2)-6(x-5)-24x=100x + y = x.y
=> xy - x - y = 0
=> (xy - x) - y + 1 = 1
=> x(y - 1) - (y - 1) = 1
=> (x - 1)(y - 1) = 1
=> x - 1 = y - 1 = 1 hoặc x - 1 = y - 1 = -1
=> x = y = 2 hoặc x = y = 0
Tim x thuoc Z
(x-3)+(x-2)+(x-1)+...+10+11=11
Giup mink cau nay nha
1,chung minh dang thuc
-(-a+b+c)+(b-c-1)=(b-c+6)-(7-a+b)+c
2,Tim x thuoc Z
(x+1)+(x+3)+(x+5)+.....+(x+99)=0
(x-3)+(x-2)+(x-1)+...+10+11=0
\(-\left(-a+b+c\right)+\left(b-c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)
\(a-b-c+b-c-1=b-c+6-7+a-b+c\)
\(a-2c-1=a-1\)
\(-2c\ne0\)hay đẳng thức ko xảy ra
tim cac so nguyen x;y sao cho
a;(x+1)+(x+3)+.....+(x+99)=100
b;(x-3)+(x-2)+(x-1)+....+10+11=11
a, (x+1) + (x+3) + ...... + (x+99) = 100
(x+x+x+...+x) + (1+3+....+99) = 100
(x.50) + 2450 = 100
x.50 = 100 - 2450
x.50 = -2350
x = -2350 : 50
x = -47
b, (x-3) + (x-2) + (x-1) + ........ + 10 + 11 = 11
(x+x+x) - (3+2+1) + (1+2+3+...+10+11) = 11
3x - 6 + 66 = 11
3x + 60 = 11
3x = 11 - 60
3x = 49
x = \(\frac{49}{3}\)
phần b mk ko chắc lắm
tim x,y thuoc Z biet (x-3)+(x-2)+(x-1)+...+10+11=11
bạn nào giải được chi tiết giúp mình thì mình tick cho nhé
tim STN x biet
a, 21 thuoc B(x-3)
b, 1-x thuoc U(17)
c, 2x+3 thuoc B(2x-1)
d, x+1 thuoc U(x mu 2+x+3)
e, 3x+1:11-2x
a: \(\Leftrightarrow x-3\inƯ\left(21\right)\)
\(\Leftrightarrow x-3\in\left\{-3;-1;1;3;7;21\right\}\)
hay \(x\in\left\{0;2;4;6;10;24\right\}\)
b: \(\Leftrightarrow x-1\in\left\{1;-1;17\right\}\)
hay \(x\in\left\{2;0;18\right\}\)
c: \(\Leftrightarrow2x-1+4⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1\right\}\)
hay \(x\in\left\{1;0\right\}\)
d: \(\Leftrightarrow x^2+x+3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)\)
\(\Leftrightarrow x+1\in\left\{1;3\right\}\)
hay \(x\in\left\{0;2\right\}\)
timx,y thuoc z biet
a,(x+1)+(x+3)+(x+5)+...+(x+99)=0 b,(x-3)+(x-2)+(x-1)+...+10+11=11
c,(x-3)+(2x+1)=7 s,(x+3).(x2+1)=0
a) (x+1)+(x+3)+(x+5)+...+(x+99)=0
=> 99x + (1 + 3 + 5 +...+ 99) = 0
Áp dụng công thức tính dãy số ta có :
1 + 3 + 5 + ... + 99 = \(\frac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}=\frac{50.100}{2}=50.50=2500\)
=> 99x + 2500 = 0
=> 99x = -2500
=> x = \(\frac{-2500}{99}=25,\left(25\right)\)
(x+1)+(x+3)+(x+5)+. . .+(x+99)=0
(x+x+. . .+x)+(1+3+5+. . .+99)=0
50x+2500=0
50x=-2500
x=-50
a) (x+1)+(x+3)+(x+5)+...+(x+99)=0
=> 99x + (1 + 3 + 5 +...+ 99) = 0
Áp dụng công thức tính dãy số ta có :
1 + 3 + 5 + ... + 99 =
2
99 − 1 :2 + 1 . 99 + 1 =
2
50.100 = 50.50 = 2500
=> 99x + 2500 = 0
=> 99x = -2500
=> x =
99
−2500 = 25, 25
ok k mk nha @@
(x+1)+(x+3)+...+(x+99)=0
(x-3)+(x+2)+(x-1)+....+10+11=11
\(\left(x+1\right)+\left(x+3\right)+.....+\left(x+99\right)=0\)
\(\left(x+x+......+x\right)+\left(1+3+........+99\right)=0\)
\(50x+2500=0\)
\(50x=-2500\)
\(x=-50\)
Tinh
1^2+2^2+...+99^2
tim x
(x-3)+(x-2)+...+10+11
1 cau =1 tick
1^2 + 2^2 + 3^2 + ... + 99^2
= 1(2 - 1) + 2(3 - 1) + 3(4- 1) + ... + 98(99-1) + 99(100 -1)
= 1.2 -1.1 + 2.3 - 2.1 + 3.4 - 3.1 + ... + 98.99 + 98.1 + 99.100 - 99
= (1.2 + 2.3 + 3.4 + ... + 99.100) - (1 + 2 + 3 + ... + 99)
đặt S = 1.2 + 2.3 + 3.4 + ... + 99.100
3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + 99.100.(101 - 98)
3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3S = 99.100.101
S = 99.100.101 : 3 = 333300
A = 1 + 2 + 3 + ... + 99
A = (99 + 1).99 : 2 = 4950
Tinh ra là :
333300 - 4950 = 328350