Tìm x biết : |x|+|x+1|+|x+2|+|x+3|+|x+4|+.......+|x+2016|+|x+2017|=1009^2
Bài 1: Tìm x,y biết:
a) 3-2*x = 3*(5-x)+4
b) 4-(7*x+2017)=6*(5-x)-2017
c) 15-x*(x+1)=4-x^2+2*x
d) -4*(x-5)+2016=3*(8-x)-(2*x-2016)
Ai nhanh mình like cho!
Tìm x biết : \(\dfrac{x+4}{2014}\)+\(\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
tìm x,y biết :
a, 3 - 2x = 3.(5-x) + 4
b, 4 - ( 7x + 2017 ) = 6 . ( 5-x) - 2017
c, 15 - x (x+1) = 4 - x^2 + 2x
d, -4.(x-5) + 2016 = 3.(8-x)-(2x - 2016)
a, 3 - 2x = 3 . (5 - x) + 4
3 - 2x = 15 - 3x + 4
-2x + 3x = 15 + 4 - 3
x = 16
b, 4 - (7x + 2017) = 6 . (5 - x) - 2017
4 - 7x - 2017 = 30 - 6x - 2017
-7x + 6x = 30 - 2017 - 4 + 2017
-x = 26
x = -26
c, 15 - x . (x + 1) = 4 - x^2 + 2x
15 - x^2 - x = 4 - x^2 + 2x
-x^2 - x + x^2 - 2x = 4 - 15
-3x = -11
x = 11/3
d, -4 . (x - 5) + 2016 = 3 . (8 - x) - (2x - 2016)
-4x + 20 + 2016 = 24 - 3x - 2x + 2016
-4x + 3x +2x = 24 + 2016 - 20 - 2016
x = 4
đúng 100%
Tìm x biết
a)x/2+x/3+x/4+x/5=0
b)x+1/2019 + x+2/2018 = x+3/2017 + x+4/2016
a) \(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}=0\)
\(\frac{77x}{60}=0\)
\(77x=0.60\)
\(77x=0\)
\(x=0\)
\(\frac{a+1}{2019}+\frac{a+2}{2018}=\frac{a+3}{2017}+\frac{a+4}{2016}\Leftrightarrow\frac{a+2020}{2019}+\frac{a+2020}{2018}=\frac{a+2020}{2017}+\frac{a+2020}{2016}\)
\(\left(a+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\Rightarrow a+2020=0\Leftrightarrow a=-2020\)
chứng tỏ số abcabc là bội cuar7,11 và 13
tìm x biết:
a) x+17=-33
b)2-(x-5)=5.2^3
c)1009.x=(-1)+2+(-3)+4+(-5)+6...9+(-2017)+2018
Ta có : \(\overline{abcabc}:\overline{abc}=1001\)
\(\Rightarrow\) \(\overline{abcabc}=\overline{abc}\times1001=\overline{abc}\times7\times11\times13\)
Vậy \(\overline{abcabc}\)là bội của 7; 11; 13
____________________________
Tìm x :
a/ \(x+17=-33\)
\(x=-33-17\)
\(x=-50\)
b/ \(2-\left(x-5\right)=5\times2^3\)
\(2-\left(x-5\right)=40\)
\(x-5=2-40\)
\(x-5=-38\)
\(x=-38+5\)
c/ \(1009.x=\left(-1\right)+2+\left(-3\right)+4+\left(-5\right)+6+....+\left(-2017\right)+2018\)
\(1009.x=\left(-1\right)+\left(-1\right)+\left(-1\right)+....+\left(-1\right)\)
\(1009.x=\left(-1\right).1010\)
\(1009.x=-1009\)
\(\Rightarrow x=-1\)
\(x=-33\)
1, ta có : abcabc=1000abc +abc=1001abc chia hết cho 7,11,13
2,a,x+17=-33 b,2-(x-5)=5.2^3 c,1009.x=-1+2+(-3)+4+...+(-2017)+2018
x=-33-17 2-(x-5)=5.8 1009.x=(-1+2)+(-3+4)+...+(-2017+2018)
x=-50 2-(x-5)=40 1009x=1+1+1+...+1+1
x-5=-38 1009.x=1009.1
x=-33 1009.x=1009
x=1
Tìm x,y biết
a)|x-3|+|7-x|=4
b)|x-2015|+|1007-1/2y|+|x-2016|+|2017-x|=2
Tìm x=?, biết:
x+(x+1)+(x+2)+(x+3)+...+(x+2016)=2017
x+(x+1)+(x+2)+(x+3)+...+(x+2016)=2017
( x + x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 2016 ) = 2017
2017x + 2033136 = 2017
2017x = 2017 - 2033136
2017x = -2031119
x = -2031119 : 2017
x = -1007
Ta có : x + (x + 1) + (x + 2) + (x + 3) +......+ (x + 2016) = 2017
=> (x + x + x + ..... + x) + (1 + 2 + 3 + .... + 2016) = 2017
=> 2017x + 2033136 = 2017
=> 2017x = 2017 - 2033136
=> 2017x = -203119
=> x = -203119 : 2017
=> x = -1007
Ta có :
x + (x + 1) + (x + 2) + (x + 3) +......+ (x + 2016) = 2017
=> (x + x + x + ..... + x) + (1 + 2 + 3 + .... + 2016) = 2017
<=> 2017x + 2 033 136 = 2017
<=> 2017x = 2017 - 2 033 136
<=> 2017x = -203 119
<=> x = -203 119 : 2017
<=> x = -1007
Tìm x, biết:
a) | x - 2017 | = 2017 - x
b) | x - 2016 | + | x - 2017 | = 2018
c) | x - 1 | + | x + 3 | = 4
Lâp bảng xét dấu
2016 2017
x-2016 _ 0 + +
x-2017 _ _ 0 +
Nếu x<2016 thì |x-2016|=2016-x,|x-2017|=2017-x
Ta có 2016-x+2017-x=2018
4033-2x=2018
2x=2015
x=1007,5
Nếu 2016<=x<=2017thif |x-2016|=x-2016;|x-2017|=2017-x
Ta có x-2016+2017-x=2018
ox+1=2018
0x=2017 (vô lí)
Nếu x>=2017 thi |x-2016|=x-2016;|x-2017|=x-2017
Ta có x-2016+x-2017=2018
2x-4033=2018
2x=6051
x=3025,5
Vậy x=1007,5 hoăc x=3025,5
Tìm x,biết:
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\)
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)
\(x+2018.\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\)\(\frac{x+1}{2017}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
=> x+2018=0
=> x=-2018