Ta có : A= 1+3+3²+...+3¹⁹⁹

Nên   : 3A= 3+3²+3³...+3²⁰⁰

Do đó: 3A-A= (3+3²+3³...+3²⁰⁰) - (1+3+3²+...+3¹⁹⁹)

                  = [ (3+3²+3³...+3¹⁹⁹) - (13+3²+...+3¹⁹⁹) ] + 3²⁰⁰ - 1

                  = 3²⁰⁰ -1

Ta lại có: 2A+1=2^x - 1

Hay       : 2 . 3²⁰⁰ - 1+1=2^x - 1

(=)         : 2 . 3²⁰⁰=2^x - 1

Vậy bài trên bị sai đề

Bài 1:

a,\(A=3+3^2+3^3+...+3^{2010}\)

\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)

\(=3\left(1+3+3^2+3^3\right)+....+3^{2007}\left(1+3+3^2+3^3\right)\)

\(=3.40+...+3^{2007}.40\)

\(=40\left(3+3^5+...+3^{2007}\right)⋮40\)

Vì A chia hết cho 40 nên chữ số tận cùng của A là 0

b,\(A=3+3^2+3^3+...+3^{2010}\)

\(3A=3^2+3^3+...+3^{2011}\)

\(3A-A=\left(3^2+3^3+...+3^{2011}\right)-\left(3+3^2+3^3+...+3^{2010}\right)\)

\(2A=3^{2011}-3\)

\(2A+3=3^{2011}\)

Vậy 2A+3 là 1 lũy thừa của 3

A = 3 + 3² + 3³ + ... + 3¹⁰⁰

⇔ 3A = 3( 3 + 3² + 3³ + ... + 3¹⁰⁰ )

⇔ 3A = 3² + 3³ + ... + 3¹⁰¹

⇔ 2A = 3A - A

          = 3² + 3³ + ... + 3¹⁰¹ - ( 3 + 3² + 3³ + ... + 3¹⁰⁰ )

          = 3² + 3³ + ... + 3¹⁰¹ - 3 - 3² - 3³ - ... - 3¹⁰⁰

          = 3¹⁰¹ - 3

2A + 3 = 3^x+100

⇔ 3¹⁰¹ - 3 + 3 = 3^x+100

⇔ 3¹⁰¹ = 3^x+100

⇔ 101 = x + 100

⇔ x = 1

Vậy x = 1

                                                        Bài giải

\(A=3+3^2+3^3+...+3^{100}\)

\(3A=3^2+3^3+3^4+...+3^{101}\)

\(3A-A=2A=3^{101}-3\)

Ta có : \(2A+3=3^{x+100}\)

\(3^{101}-3+3=3^{x+100}\)

\(3^{101}=3^{x+100}\)

\(\Rightarrow\text{ }x+100=101\)

\(\Rightarrow\text{ }x=1\)

A=3+3²+3³+...+3²⁰¹⁹

3A=3²+3³+...+3²⁰²⁰

3A-A=(3²+3³+...+3²⁰²⁰)-(3+3²+3³+...+3²⁰¹⁹)

2A=3²⁰²⁰-3

2A+3=3²⁰²⁰

⇒n=2020

trả lời câu c nha

A=3+3^2 +3^+...+3^99+3^100

3A=3^2+3^3+...+3^100+3^101

3A-A=2A=3^101-3

Do đó 2A+3=3^101.Theo đề bài,2A+3=3^x

Vậy x=101

^ là mụ nha

\(a,\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)

\(=\left(2a+3\right)\left(x-y+1\right)\)

\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

​\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)​

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(c,x^k+1-x^k-1\)

\(=0?!?!\)

\(d,x^m+3-x^m+1\)

\(=4\)

\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)

\(=\left(x-y\right)^2\left(3\left(x-y\right)-2\right)\)

\(=\left(x-y\right)^2\left(3x-3y-2\right)\)

\(f,81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a+1\)

\(=\left(9a+1\right)^2\)

\(g,25a^2.b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

\(h,\left(a-b\right)^2-2\left(a-b\right)c+c^2\)

\(=\left(a-b-c\right)^2\)

\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)

\(=2by.2ax\)

\(=4axby\)

\(\text{a) }\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)

\(=\left(2a+3\right)\left(x-y+1\right)\)

\(\text{b) }\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(\text{c) }x^k+1-x^k-1\)

\(=\left(x^k-x^k\right)+\left(1-1\right)\)

\(=0\)

\(\text{d) }x^m+3-x^m+1\)

\(=\left(x^m-x^m\right)+\left(3+1\right)\)

\(=4\)

\(\text{e) }3\left(x-y\right)^3-2\left(x-y\right)^2\)

\(=\left(x-y\right)^2\left[3\left(x-y\right)-2\right]\)

\(=\left(x-y\right)^2\left(3x-3y-2\right)\)

\(\text{f) }81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a.1+1^2\)

\(=\left(9a+1\right)^2\)

\(\text{g) }25a^2b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

\(\text{h) }\left(a-b\right)^2-2.\left(a-b\right).c+c^2\)

\(=\left(a-b-c\right)^2\)

\(\text{i) }\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by+ax-by\right)\left(ax+by-ax+by\right)\)

\(=2by.2ax\)

\(=4byax\)

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

3. \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3+48x^2-12x+9\)

\(=8\)