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3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
a, A = 1 + 3 + 32 + 33 +....+32022
3A = 3 + 32 + 33 +.....+32022 + 32023
3A - A = 32023 - 1
2A = 32023 - 1
2A - 22023 = 32023 - 1 - 22023
2A - 22023 = -1
b, x \(\in\) Z và x + 10 \(⋮\) x - 1 ( đk x# 1)
x + 10 \(⋮\) x - 1
\(\Leftrightarrow\) x - 1 + 11 \(⋮\) x - 1
11 \(⋮\) x - 1
x-1 \(\in\) { -11; -1; 1; 11}
x \(\in\) { -10; 0; 2; 12}
Kết luận các số nguyên x thỏa mãn yêu cầu đề bài là :
x \(\in\) { -10; 0; 2; 12}
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
a,
A = 1 + 3 + 32 + 33 + ... + 3119
3A = 3.(1 + 3 + 32 + 33 + ... + 3119)
3A = 3 + 32 + 33 + 34+ ... + 3120
2A = 3A - A = (3 + 32 + 33 + 34 + ... + 3120) - (1 + 3 + 32 + 33 + ... + 3119)
2A = 3120 - 1
A = \(\frac{3^{120}-1}{2}\)
Vậy A = \(\frac{3^{120}-1}{2}\)
b, Ta có : 3120 - 1 + 1 = 27x
<=> 3120 = 27x
<=> 3120 = (33)x
<=> 3120 = 3x
<=> x = 120
Vậy x = 120
c, A có chia hết cho 5 và 13
Sua cho \(\left(3^3\right)^x=3^{3x}\) nha
\(\Rightarrow3^{120}=3^{3x}\Rightarrow x=\frac{120}{3}=40\)
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
