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3x + ( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ... + ( x + 99 ) = 2765
= 3x + ( x + x + x + ... + x ) + ( 1 + 3 + 5 + ... + 99 ) = 2765
1 + 3 + 5 + ... + 99 = ( 99 + 1 ) . 50 : 2 = 2500 ( Ở đây mình không tiện tính số số hạng )
=> 3x + ( x + x + x + ... + x ) = 2765 - 2500
3x + ( x + x + x + ... + x ) = 265
3x + ( x + x + x + ... ) = 53x = 265
=> x = 265 : 53
x = 5
\(3x+\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=2765\)
\(\Leftrightarrow3x+\left(x+x+x+...+x\right)+\left(1+3+5+...+99\right)=2765\)(50 số hạng x)
\(\Leftrightarrow3x+50x+2500=2760\Leftrightarrow3x+50x=2765-2500\)
\(\Leftrightarrow53x=265\Leftrightarrow x=\frac{265}{53}=5\)
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\(3x+\left(x+1\right)+\left(x+3\right)+...........+\left(x+99\right)=2765\)
\(\Leftrightarrow3x+\left(x+x+.....+x\right)+\left(1+3+....+99\right)=2759\)
\(\Leftrightarrow3x+50x+2500=2759\)
\(\Leftrightarrow53x+2500=2765\)
\(\Leftrightarrow53x=265\)
\(\Leftrightarrow x=5\)
Vậy .....
3x + (x + 1) + (x + 3) + (x + 5) + ... + (x + 99) = 2765
3x + 50(x + 1 + x + 99) : 2 = 2765
3x + 25(x . 2 + 100) = 2765
3x + 50x + 2500 = 2765
53x = 265
x = 5
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a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555
=> 101x +5050 = 5555
=> 101x = 505
=> x = 505 : 101 = 5
Vậy, x = 5
b)1+2+3+4+...+x=820
=> ( x+1) x :2 = 820
=> (x+1)x = 1640
Mà 1640 = 40 . 41
=> x = 40 ( vì {x+1} - x = 1)
Vậy, x = 40
c) 3x+1 = 9.27=243
=> 3x+1 = 35
=>x + 1 = 5
=> x = 4
Vậy, x=4
d) x+2x+3x+...+99x+100x=15150
=> [( 100 + 1) x 100 :2 ] x = 15150
=> 5050x = 15150
=> x = 15150:5050 = 3
Vậy, x =3
e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050= 200500
=> x = 200500 : 100 = 2005
Vậy, x = 2005
f)3x+3x+1+3x+2=351
=> 3x + 3x . 3 + 3x x 9 = 351
=> 3x ( 1+3+9) = 351
=> 3x . 13 = 351
=> 3x = 351 :13=27 mà 27 = 33
=> x=3
Vậy, x=3
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(x+1)+(2x+3)+(3x+5)+...+(50x+99)=3775
=> (x + 2x + 3x + ... + 50x) + (1 + 3 + 5 + ... + 99) = 3775
=> x . (1 + 2 +3 + ...+ 50) + (1 + 3 + 5 + ... + 99) = 3775
Áp dụng công thức tính dãy số ta có :
\(1+2+3+...+50=\frac{\left[\left(50-1\right):1+1\right].\left(50+1\right)}{2}=\frac{50.51}{2}=25.51=1275\)
\(1+3+5+...+99=\frac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}=\frac{50.100}{2}=50.50=2500\)
=> x . 1275 + 2500 = 3775
=> 1275x = 1275
=> x = 1
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a: \(\left(x+1\right)+\left(x+3\right)+...+\left(x+99\right)=0\)
\(\Leftrightarrow50x+2500=0\)
=>50x=-2500
hay x=-50
c: (x-7)(x+3)<0
=>x+3>0 và x-7<0
=>-3<x<7
b: (x-3)(2y+1)=7
\(\Leftrightarrow\left(x-3;2y+1\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(4;3\right);\left(10;0\right);\left(2;-4\right);\left(-4;-1\right)\right\}\)
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3(x + 6) = 2(x - 5)
=> 3x + 18 = 2x - 10
=> 3x - 2x = -10 - 18
=> x = -28
vay_
c, |3x - 7| = 6
=> 3x - 7 = 6 hoac 3x - 7 = -6
=> 3x = 13 hoac 3x = 1
=> x = 13/3 hoac x = 1/3
vay_
3(x + 6) = 2(x - 5)
=> 3x + 18 = 2x - 10
=> 3x - 2x = -10 - 18
=> x = -28
Vậy x = -28
c, |3x - 7| = 6
=> 3x - 7 = 6 hoac 3x - 7 = - 6
=> 3x = 13 hoac 3x = 1
=> x = \(\frac{13}{3}\) hoac x = \(\frac{1}{3}\)
Vậy \(x\in\left\{\frac{3}{13};\frac{1}{3}\right\}\)
3x + (x+1) + (x+3)+(x+5)+...+(x+99) = 2765
3x + 50.x +[(1+99).50:2] = 2765
53.x + 2500 = 2765
53x = 265
x = 5