tìm x
[x+1+x+2+x+3+.......+x+100]= 5950
giải gấp giúp mình
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( 1 + x ) + ( 2 + x ) + ( 3 + x ) + ...+ ( 100 + x ) = 2018
\(\Rightarrow\)( 1 + 2 + 3 + ... + 100 ) + ( x + x + x + ... + x ) = 2018
\(\Rightarrow\){( 1 + 100 ) . [( 100 - 1 ) : 1 + 1 ] : 2 } + ( x + x + ... + x ) = 2018
\(\Rightarrow\)5050 + x . 100 = 2018
\(\Rightarrow\) x100 = 2018 - 5050 = -3032
\(\Rightarrow\)x = -3032 : 100 = -30,32
vậy x = -30,32
mà nè sai thì xin lỗi đề hơi có vấn đề nếu sai thì sorry nha !!!
ta có 605x > 0 suy ra x>0
ta có x+1+x+2+x+...+100=605x
100x +5050=605x
505x=5050
x=10
\(6\cdot x-5=613\)
\(6\cdot x=613+5\)
\(6\cdot x=618\)
\(x=618\div6\)
\(x=103\)
Vậy \(x=103\)
\(12\cdot x+3\cdot x=30\)
\(x\cdot\left(12+3\right)=30\)
\(x\cdot15=30\)
\(x=30\div15\)
\(x=2\)
Vậy \(x=2\)
\(125-25\cdot\left(x-1\right)=100\)
\(25\cdot\left(x-1\right)=125-100\)
\(25\cdot\left(x-1\right)=25\)
\(x-1=25\div25\)
\(x-1=1\)
\(x=1+1\)
\(x=2\)
Vậy \(x=2\)
\(\left(x-2\right)\cdot\left(x-14\right)=0\)
\(\Rightarrow\) \(x-2=0\) hoặc \(x-14=0\)
TH1: \(x-2=0\) TH2: \(x-14=0\)
\(x=0+2\) \(x=0+14\)
\(x=2\) \(x=14\)
Vậy \(x=2\) hoặc \(x=14\)
\(128-3\cdot\left(x+4\right)=23\)
\(3\cdot\left(x+4\right)=128-23\)
\(3\cdot\left(x+4\right)=105\)
\(x+4=105\div3\)
\(x+4=35\)
\(x=35-4\)
\(x=31\)
Vậy \(x=31\)
12.x+3.x=30
x.(12+3)=30
x.15=30
x =30:15
x =2
125-25.(x-1)=100
25.(x-1)=125-100
25.(x-1)=25
x-1=25:25
x-1=1
x =1+1
x=2
(x-2).(x-14)=0
x=14
128-3.(x+4)=23
3.(x+4)=128-23
3.(x+4)=105
x+4=105:3
x+4=35
x = 35+4
x =39
a,2x+53=135
2x=135-53
2x=82
x=82:2
x=41
bạn viết khó hỉu quá nên mk giúp bạn dc câu a thui
k hộ mk với
\(A=3^x+3^{x+1}+...+3^{x+100}\)
=>\(3A=3^{x+1}+3^{x+2}+...+3^{x+101}\)
=>\(2A=3^{x+101}-3^x\)
=>\(A=\dfrac{3^{x+101}-3^x}{2}\)
=>\(3^{x+101}-3^x=3^{105}-3^4\)
=>x=4
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
(x + 1 ) + (x + 2) + (x + 3) + ... + (x + 100) = 7450
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 7450
100 số x 100 số hạng
100.x + (1 + 100).100:2 = 7450
100.x + 101.50 = 7450
100.x + 5050 = 7450
100.x = 7450 - 5050
100.x = 2400
x = 2400 : 100
x = 24
Vậy x = 24
(x+1)+(x+2)+(x+3)+......+(x+100)=7450
=> 100x + (1 + 2 + 3 + ...+ 100) = 7450
=> 100x + 5050 = 7450
=> 100x = 2400
=> x = 24
(x - 1) (x + 1) - (x + 1) (x + 2) = - 19
<=> x² - 1 - (x² + 3x + 2) = - 19
<=> x² - 1 - x² - 3x - 2 = -19
<=> - 3x = -16
<=> x = 16/3
vậy x = 16/3
(2x - 1) .3 - (x + 3). (-2) = -x +1
<=> (6x - 3) - (-2x - 6 ) = 1 - x
<=> 6x - 3 + 2x + 6 = 1 - x
<=> 8x + 3 = 1 - x
<=> 9x = - 2
<=> x = - 2/9
vậy x = - 2/9
Bài 1: Tìm x
a) x . (x + 3) = 0
=> \(\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=0-3\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
b) (x -1) (x2 - 1) = 0
=> \(\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0+1\\x^2=0+1\left(bỏ\right)\end{cases}}\)
=> x = 1
Bài 2: Tìm x, biết
a) -12(x - 5) + 7(3 - x) = 5
-12x - (-12 . 5) + 7 . 3 - 7x = 5
-12x + 60 + 21 - 7x = 5
-12x - 7x = 5 - 21 - 60
-19x = -76
x = -76 : (-19)
x = 4
\(\left[x+1+x+2+x+3+...+x+100\right]=5950\)
\(\left[\left(x+x+x+...+x\right)+\left(1+2+3+4+...+100\right)\right]=5950\)
\(\left[\left(x+x+x+...+x\right)+\left(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\right)\right]=5950\)
\(\left[100x+5050\right]=5950\)
\(100x=900\Leftrightarrow x=9\)