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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
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x + 1) = 99/100
1- 1/2 +1/2-1/3+1/3-1/4+...+ 1/x - 1/ x+ 1 = 99/100
1 - 1/ x+1 = 99/ 100
=> (100 - 1)/ x+1 = 99 / 100
=> x+1 = 100 => x=99
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
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
cho \(M=1+3+3^2+...+3^{99}+3^{100}\)
=>\(M=1+\left(3+3^2+3^3\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=>M=1+3\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=>M=1+13\left(3+...+3^{98}\right)\)
Mà \(13\left(3+3^{98}\right)⋮13\)
=> M chia cho 13 dư 1
+) \(M=1+3+3^2+...+3^{99}+3^{100}\)
\(\Leftrightarrow M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(\Leftrightarrow M=\left(1+3+9\right)+3^3\left(1+3+9\right)+....+3^{98}\left(1+3+9\right)\)
\(\Leftrightarrow M=13+3^3\cdot14+....+3^{98}\cdot14\)
\(\Leftrightarrow M=13\left(1+3^3+....+3^{98}\right)\)
=> M chia 13 dư 0
\(A=20\times21+21\times22+...+99\times100\)
\(3\times A=20\times21\times\left(22-19\right)+21\times22\times\left(23-20\right)+...+99\times100\times\left(101-98\right)\)
\(=20\times21\times22-19\times20\times21+...+99\times100\times101-98\times99\times100\)
\(=99\times100\times101-19\times20\times21\)
Suy ra \(A=\frac{99\times100\times101-19\times20\times21}{3}=360640\)
\(B=3\times4\times5+4\times5\times6+...+98\times99\times100\)
\(4\times B=3\times4\times5\times\left(6-2\right)+4\times5\times6\times\left(7-3\right)+...+98\times99\times100\times\left(101-97\right)\)
\(=3\times4\times5\times6-2\times3\times4\times5+...+98\times99\times100\times101-97\times98\times99\times100\)
\(=98\times99\times100\times101-2\times3\times4\times5\)
Suy ra \(B=\frac{98\times99\times100\times101-2\times3\times4\times5}{4}=24497520\)
x=(1+2+3-4-5-6)+...+(97+98+99-100-101-102)
x=-9+...+-9
x=-9.17
x=-153