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x+2x+3x+...+100x=5500
5500=((1+100)*50)x
=5050
bạn sai đề

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

\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)
\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)
\(100x-5050=50\)
\(100x=50+5050\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=7\)
\(b,x+\left(1+2+3+...+50\right)=2000\)
\(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)
\(x+1275=2000\)
\(\Rightarrow x=2000-1275=725\)

Ta thấy:
1 x 4 = 1 x 2 + 1 x 2
2 x 5 = 2 x 3 + 2 x 2
3 x 6 = 3 x 4 + 3 x 2
.................................
Suy ra:
D = (1 x 2 + 2 x 3 + 3 x 4 + .... + 97 x 98) + (1 x 2 + 2 x 2 + 3 x 2 + .... + 97 x 2)
D = (1x2+2x3+3x4+...+97x98) + (1+2+3+...+99)x2
D = (1x2+2x3+3x4+...+97x98) + 100 x 99 : 2
D - 100 x 99 : 2 = 1x2+2x3+3x4+...+97x98
D - 4950 = 1x2+2x3+3x4+...+97x98
(D - 4950) x 3 = 1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+......+97x98x(99-96)
(D-4950)x3 = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 97 x 98 x 99 - 96 x 97 x 98
(D-4950)x3 = 97 x 98 x 99
Và từ đây ta có thể tìm hướng để ra kết quả

a) Số số hạng: \(\frac{\left(99-1\right)}{1}+1=99\)
Tổng: \(\frac{99+1}{2}\cdot99=4950\)
b) Số số hạng: \(\frac{\left(100-2\right)}{2}+1=50\)
Tổng: \(\frac{100+2}{2}\cdot50=2550\)
c) \(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(3\cdot S=1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+3\cdot4\left(5-2\right)+...+99\cdot100\left(101-98\right)\)
\(3\cdot S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3\cdot S=99\cdot100\cdot101\)
Vậy, \(S=\frac{1}{3}\cdot99\cdot100\cdot101=333300\)

a) \(2^{x+2}-2^2=96\)
<=> \(2^x.2^2-2^x=96\)
<=> \(2^x\left(4-1\right)=96\)
<=> \(3.2^x=96\)
<=> \(2^x=32\)
<=> \(2^x=2^5\)
<=> x = 5
b, \(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\)
\(\Rightarrow x-\left(\frac{1x}{2}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\left(\frac{4x}{8}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{8x}{8}-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\Rightarrow x=\frac{45}{4}\div\frac{3}{8}=30\)
Vậy x = 30

\(a,2^x.100=100^4\) \(b,100^3:2^x=100\)
\(\Rightarrow2^x=100^4:100\) \(\Rightarrow2^x=100^3:100\)
\(\Rightarrow2^x=100^3\) \(\Rightarrow2^x=100^2\)
\(\Rightarrow x=\sqrt{100^3}\) \(\Rightarrow x=\sqrt{100^2}\)
\(\Rightarrow x=1000\) \(\Rightarrow x=100\)
VẬY X=1000 VẬY X=100
\(\left(x+2\right)+\left(x+4\right)+...+\left(x+100\right)=5500\)
\(\Rightarrow\left(x+x+...+x\right)\) (có 50 số x) \(+\left(2+4+...+100\right)=5500\)
Số hạng là:
\(\left(100-2\right):2+1=50\) (số hạng)
Tổng là:
\(\left(100+2\right)\cdot50:2=2550\)
\(\Rightarrow50x+2550=5500\\ \Rightarrow50x=5500-2550\\ \Rightarrow50x=2950\\ \Rightarrow x=59\)