`#3107`

\(A=1+2^1+2^2+2^3+...+2^{2015}\)

\(2A=2+2^2+2^3+2^4+...+2^{2016}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)

\(A=2+2^2+2^3+2^4+...+2^{2016}-1-2-2^2-2^3-...-2^{2015}\)

\(A=2^{2016}-1\)

Vậy, \(A=2^{2016}-1.\)

\(A=2^0+2^1+2^2+...+2^{2015}\)

\(2\cdot A=2^1+2^2+2^3+...+2^{2016}\)

\(A=2A-A=2^{2016}-2^0\)

\(A=2^{2016}-1\)

A=(1+2+2^2)+2^3(1+2+2^2)+...+2^2013(1+2+2^2)+2^2016

=7(1+2^3+...+2^2013)+2^2016

Vì 2^2016 chia 7 dư 1

nên A chia 7 dư 1

Ta có: \(A=1+2+2^2+...+2^{2015}\)

\(2A=2\cdot\left(1+2+2^2+...+2^{2015}\right)\)

\(2A=2+2^2+2^3+...+2^{2016}\)

\(2A-A=2+2^2+...+2^{2016}-1-2-2^2-...-2^{2015}\)

\(A=2^{2016}-1\)

A không thể biết dưới dạng lũy thừa của 8 được 

A=22016−1

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:

\(8^x-1=2^{2016}-1\)

\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)

\(\Rightarrow2^{3x}-1=2^{2016}-1\)

\(\Rightarrow2^{3x}=2^{2016}\)

\(\Rightarrow3x=2016\)

\(\Rightarrow x=\dfrac{2016}{3}\)

\(\Rightarrow x=672\)

Bài 1

S₂ = 21 + 23 + 25 + ... + 1001

Số số hạng của S₂:

(1001 - 21) : 2 + 1 = 491

⇒ S₂  = (1001 + 21) . 491 : 2 = 250901

--------

S₄  = 15 + 25 + 35 + ... + 115

Số số hạng của S₄:

(115 - 15) : 10 + 1 = 11

⇒ S₄ = (115 + 15) . 11 : 2 = 715

Bài 2

a) 2x - 138 = 2³.3²

2x - 138 = 8.9

2x - 138 = 72

2x = 72 + 138

2x = 210

x = 210 : 2

x = 105

b) 5.(x + 35) = 515

x + 35 = 515 : 5

x + 35 = 103

x = 103 - 35

x = 78

c) 814 - (x - 305) = 712

x - 305 = 814 - 712

x - 305 = 102

x = 102 + 305

x = 407

d) 20 - [7.(x - 3) + 4] = 2

7(x - 3) + 4 = 20 - 2

7(x - 3) + 4 = 18

7(x - 3) = 18 - 4

7(x - 3) = 14

x - 3 = 14 : 7

x - 3 = 2

x = 2 + 3

x = 5

e) 9ˣ⁻¹ = 9

x - 1 = 1

x = 1 + 1

x = 2

a: \(280-\left(x-140\right):35=270\)

=>(x-140):35=280-270=10

=>x-140=350

=>x=350+140

=>x=490

b: \(\left(190-2x\right):35-32=16\)

=>\(\left(190-2x\right):35=32+16=48\)

=>\(190-2x=35\cdot48=1680\)

=>2x=190-1680=-1490

=>x=-745

c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)

=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)

=>41-(2x-5)=720:40=18

=>2x-5=41-18=23

=>2x=28

=>x=14

d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)

=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)

=>\(\frac{x}{23}+45=46\)

=>\(\frac{x}{23}=1\)

=>x=23

e: \(\left(3x-4\right)\left(x-1\right)^3=0\)

=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)

f: \(2^{2x-1}:4=8^3\)

=>\(2^{2x-1-2}=2^9\)

=>2x-3=9

=>2x=12

=>x=6

g: \(x^{17}=x\)

=>\(x^{17}-x=0\)

=>\(x\left(x^{16}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

h: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)

=>x∈{4;5;6}

i: \(\left(x+2\right)^5=2^{10}\)

=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)

=>x+2=4

=>x=2

k: 1+2+3+...+x=78

=>\(\frac{x\left(x+1\right)}{2}=78\)

=>x(x+1)=156

=>\(x^2+x-156=0\)

=>(x+13)(x-12)=0

=>x=-13(loại) hoặc x=12(nhận)

l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

n: \(5^{x}:5^2=125\)

=>\(5^{x-2}=5^3\)

=>x-2=3

=>x=5

m: \(\left(x+1\right)^2=\left(x+1\right)^0\)

=>\(\left(x+1\right)^2=1\)

=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)

o: Số số hạng của dãy số 2;4;..;52 là:

(52-2):2+1=50:2+1=25+1=26(số)

Tổng của dãy số 2;4;...;52 là:

\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)

(2+x)+(4+x)+...+(52+x)=780

=>26x+702=780

=>26x=78

=>x=3

p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)

=>ƯCLN(70;80)=\(2\cdot5=10\)

70⋮x; 80⋮x

=>x∈ƯC(70;80)

=>x∈Ư(10)

mà x>8

nên x=10

q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)

=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)

x⋮12; x⋮25; x⋮30

=>x∈BC(12;25;30)

=>x∈B(300)

mà 0<x<500

nên x=300

\(A=1+2^1+2^2+...+2^{2015}\)

\(2\cdot A=2^1+2^2+2^3+...+2^{2015}+2^{2016}\)

\(2A-A=2^1+2^2+2^3+...+2^{2015}+2^{2016}-\left(1+2^1+2^2+...+2^{2015}\right)\)

\(A=2^{2016}-1\)

B

`=>` `B`

Vì các phân số cồn lại thuộc dạng `(x >= 1)`

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18