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b/100x+(1+2+3+...+100)=205550
100x+5050=205550
100x=205550-5050
100x=200500
x=200500/100
x=2005
d/(3x-24).75=2.76.1/20090
(3x-24).75=2.76.1
(3x-24)=2.76:75
(3x-24)=2.7
3x-16=14
3x=14+16
3x=30
x=30/10=3
Tìm x, biết:
a) (x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 205550
b) 3x + 3x + 1 + 3x + 2 = 351
a)(x+x+x+...+x)+(1+2+3+...+100)=205550
100x + 5050 =205550
100x =205550 - 5050
100x =200500
x =200500 : 100
x =2005
(x+1)+(x+2)+................+(x+100)=205550
=>x+1+x+2+................+x+100=205550
=>100x+(1+2+3+...........+100)=205550
=>100x+5050=205550
=>100x=205550-5050
=>100x=200500
=>x=200500:100
=>x=2005
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + .... + ( x + 100 ) = 205550
=> x + 1 + x + 2 + x + 3 + .... + x + 100 = 205550
=> ( x + x + x + .... + x ) + ( 1 + 2 + 3 + .... + 100 ) = 205550
=> 100x + { [ 100 . ( 100 + 1 ) ] : 2 } = 205550
=> 100x + [( 100 . 101 ) : 2 ] = 205550
=> 100x + ( 10100 : 2 ) = 205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050
=> 100x = 200500
=> x = 200500 : 100
=> x = 2005
Vậy x = 2005
sửa :
Số các số tự nhiên từ 1 đến 100 là:
[100-1] : 1 +1 =100 [số]
(x-1)+(x-2)+(x-3)+(x-4)+...+(x-100)=205550
=> 100.x -[1+2+3+..+100] = 205550
=> 100.x - 100x101 : 2 = 205550
=> 100.x -5050= 205550
=> 100.x = 210600 => x= 2106
a, => 3x-17 = 0 hoặc 3x-17 = 1
=> x=17/3 hoặc x=6
b, => x+1+x+2+....+x+100=205550
=>100x + (1+2+...+100)=205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050 = 200500
=>x= 2005
c,=>x+x+1+....+x+2010=2029099
=>2011x+(1+2+....+2010)=2029099
=>2011x+2021055=2029099
=>2011x = 2029099-2021055 = 8044
=>x=4
Có : 3Q = 3+3^2+....+3^101
2Q=3Q-Q= (3+3^2+....+3^101)-(1+3+3^2+...+3^100) = 3^101-1
=>Q = (3^101-1)/2
Bài 1.
a)Có
b)Không
Bài 2.
bỏ qua
Bài 3.
a) bỏ qua
b) 1212
(x+1)+(x+2)+(x+3)+...+(x+100)=205550
x+1+x+2+x+3+...+x+100=205550
100x+(100+1).100:2=205550
100x+5050=205550
100x=200500
x=2005
a) (x+1)+(x+2)+(x+3)+.....+(x+100)=205550
\(\left(\frac{100-1}{1}\right)+1\)=100(ngoặc)
100X+(1+2+3+.....+100)=205550
100X+5050=205550
100X=205550-5050
100X=200500
X=2005
còn lại tự làm và thêm văn võ chut ít vào đó nhé!
Bài 1:tìm x thuộc Z
a)x.(x-1)=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy: \(x=0;1\)
b)(x-3).(x+4)=0
\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Vậy: \(x=3;-4\)
c)(2x-4).(x+2)=0
\(\Leftrightarrow2\left(x-2\right).\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x=2;-2\)
d)(x+1)^2.(x-2)^2=0
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x=-1;2\)
e) x(x+1).(x+2)^2.(x+3)^3=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy: \(x=0;-1;-2;-3\)
f)(x-9)^5.(x-5)^8=0
\(\Leftrightarrow\left[\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=9\\x=5\end{matrix}\right.\)
Vậy: \(x=9;5\)
g)x(x+100)^10.(x+2000)^20.(x+300)^300=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x+100=0\\x+200=0\\x+300=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-100\\x=-200\\x=-300\end{matrix}\right.\)
Vậy: \(x=0;-100;-200;-300\)
h)(x-2)^2=0
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy: \(x=2\)
A) x.(10-5)= 0 : 2
x.5=0
x=0 : 5
x=0
B) x.101 + (1+2+3+.......+100)=205550
x.101+2525=205550
X.101 = 203025
X = 203025 : 101
a) (x-5).(2x-10)=0
TH1:x-5=0 x=5
TH2:2x-10=0 2x=10 x=5
b)(x+x+...+x) + (1+2+3+4+...+100)=205550
100x + (1+100)+(2+99)+(3+98)+....=205550
100x + 101.50 = 205550
100x + 5050 = 205550
100x = 200500
x=2005
Cho mình 1 k nha!