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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
a: =>4/3x=1/2
hay x=1/2:4/3=3/8
b: =>-3(2-3x)=4(x+1)
=>-6+9x=4x+4
=>5x=10
hay x=2
a: =>4/3x=1/2
hay x=1/2:4/3=3/8
b: =>-3(2-3x)=4(x+1)
=>-6+9x=4x+4
=>5x=10
hay x=2
\(\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x^3-6\text{x}^2+12\text{x}-8\right)+\left(9\text{x}^2-1\right)-\left(x^3+3\text{x}^2+3\text{x}+1\right)=0\)
\(\Leftrightarrow x^3-6\text{x}^2+12\text{x}-8+9\text{x}^2-1-x^3-3\text{x}^2-3\text{x}-1=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-6\text{x}^2+9\text{x}^2-3\text{x}^2\right)+\left(12\text{x}-3\text{x}\right)+\left(-8-1-1\right)=0\)
\(\Leftrightarrow9\text{x}-10=0\)
\(\Leftrightarrow9\text{x}=10\Leftrightarrow x=\frac{10}{9}\)
Vậy x = \(\frac{10}{9}\)
( 2. l x l - 1) .(7 - 3x ) =0 ( x2 + 1 ). ( 1/2 - 3x ) <0 ( l x l + 1 ) . ( 15x - 1 ) = 0
=> 2 . l x l - 1 = 0 hoặc 7 - 3x = 0 => x2+1 hoặc 1/2 -3x < 0 => l x l + 1 hoặc 15x - 1 =0
+ 2 . l x l - 1 = 0 => 2 . l x l =1 => x = 1/2 + x2 +1< 0 => x không tồn tại + l x l - 1 = 0 => l x l = 1 => thuộc 1 : -1
+ 7 - 3x = 0 => 3x = 7 => x = 7/3 + 1/2 - 3x < 0 => 3x > 1/2 => x > 1 + 15x - 1 = 0 => 15x =1 => x = 1/15
5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
nhìn cái đề con hơi bị ''sốc'' , thế này ạ ???
Sửa đề \(4+\frac{1}{3}x\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}x\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(4+\frac{1}{3}x\left(-\frac{1}{3}\right)\le x\le\frac{2}{3}x\left(-\frac{11}{12}\right)\)
\(4-\frac{1}{9}x\le x\le-\frac{11}{18}x\)
b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-7;-9;-3;-13\right\}\)
[1-3x]=2-x<=>1-3x+x-2=0<=>-2x-1=0<=>-2x=1<=>x=-1/2