( 1/2 x 2/1 - 2/3 x 3/2 + 3/4 x 4/3 - 100/101 x 101/100 + 55 ) x x = aa
( 1/2 x 2/1 - 2/3 x 3/2 + 3/4 x 4/3 - 100/101 x 101/100 + 55 ) x a = a x aa
tìm a và aa
1 x 2 + 2 x 3 + 3 x 4 + ... + 100 x 101 =
1 x 2 x 3 + 2 x 3 x 4 + ... + 100 x 101 x 102
Có cả lời giải nhé
a: S=1(1+1)+2(1+2)+...+100(1+100)
=1+2+...+100+1^2+2^2+...+100^2
\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)
=343450
b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)
=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)
=>4*A=100*101*102*103
=>A=25*101*102*103
Tính
a) (x-1/2)+(x-1/4)+(x-1/8)+...+(x-1/512)
Tìm x
a) (x-1/1×2)+(x-1/2×3)+...+(x-1/100×101)
b) (x-1)+(x-2)+(x-3)+...+(x-101)=5050
c) x+1/2+1/3+1/4+...+1/100=3/2+4/3+5/4++...+101/100
Câu 2:
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
Có \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge0\)
do đó phương trình ban đầu tương đương với:
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(\Leftrightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
Cho A = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 100 x 101
và B = 1 x 3 + 2 x 4 + 3 x 5 + 4 x 6 + ... + 100 x 102
Vậy B - A = ?
Ta có : A = 1.2 + 2.3 + 3.4 + ...... + 100.101
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 100.101.102
=> 3A = 100.101.102
=> A = 100.101.102/3
=> A = 343400
Tính tổng :
S = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 99 x 100 + 100 x 101
TL :
= 3 333 000
_HT_
S= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
S x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300
TL
=3333000
Hok tốt nhe bn
#Kirito
Cho ham so
f(x)=4^x/4^x+2
Tinh A=f(0)+f(1/101)+f(2/101)+f(3/101)+...+f(100/101)+f(1)
A=f(0)+(f(1/101)+f(100/101))+(f(2/101)+f(99/101))+...+f(1)
A=f(0)+50f(1)+f(1)
A=f(0)+51f(1)
A=4^0/4^0+2+51(4^1/4^1+2)
A=1/3+34
A=103/3
Mik ko bik đúng ko nữa
1 x 3+2 x 4+3 x 5+4 x 6+...+99 x 101+100 x 102
Cho x1+x2+x3+...+x100+x101=0. Biết x1+x2=x3+x4=...=x99+x100=x100+x101=1. Tính x100
Tính x1 + x2 +...+ x99 + x100 + x101 = 0
(x1 + x2)+ ...+ ( x99 + x100)+ x101 = 0
1 + ... + 1 + x101 = 0
1 x 50 + x101 = 0
50 + x101 = 0
x101 = 0 - 50
x101 = -50
Ta có: x100 + x101 = 1
x100 + (-50) = 1
x100 = 1-(-50)
x100 =51
Vậy x101 = 51