Chihuahua

Ta có:

\(B\left(2\right)=\left\lbrace0;\pm2;\pm4;\pm6;\pm8;\ldots\right\rbrace\)

Mà x ∈ N*, x < 6

⇒ x ∈ {2; 4}

Vậy x ∈ {2; 4}

a) Ta có:

\(C=1+4+4^2+4^3+4^4+4^5+4^6\)

\(4C=\left(1+4+4^2+4^3+4^4+4^5+4^6\right)\cdot4\)

\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)

b)

\(\Rightarrow3C=4C-C=\left(4+4^2+4^3+4^4+4^5+4^6+4^7\right)-\left(1+4+4^2+4^3+4^4+4^5+4^6\right)\)

\(3C=4^7-1\)

Vì \(3C\vdots3\) nên \(4^7-1\vdots3\)

Vậy ....

a) Ta có:

\(A=1+3+3^2+3^3+3^4+3^5+3^6+3^7\)

\(3A=\left(1+3+3^2+3^3+3^4+3^5+3^6+3^7\right)\cdot3\)

\(3A=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\)

b)

\(\Rightarrow2A=3A-A=\left(3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\right)-\left(1+3+3^2+3^3+3^4+3^5+3^6+3^7\right)\)

\(2A=3^8-1\)

Mà \(2A\vdots2\)

\(\Rightarrow3^8-1\vdots2\)

Vậy ...

SAO V BN?

Ta có:

\(A=1+2^1+2^2+\cdots+2^{2007}\)

\(2A=2+2^2+2^3+\cdots+2^{2008}\)

\(2A-A=\left(2+2^2+2^3+\cdots+2^{2008}\right)-\left(1+2^1+2^2+\cdots+2^{2007}\right)\)

\(A=2^{2008}-1\)

Vậy:

\(a)2A=2+2^2+2^3+\cdots+2^{2008}\)

\(b)A=2^{2008}-1\)

???????????????

\(\left(6-x\right)^3=\frac{-125}{8}\)

\(\left(6x-3\right)^3=\left(\frac{-5}{2}\right)^3\)

\(6x-3=\frac{-5}{2}\)

\(6x=\frac{-5}{2}+3\)

\(6x=\frac{-5}{2}+\frac62\)

\(6x=\frac12\)

\(x=\frac12:6\)

\(x=\frac12\cdot\frac16\)

\(x=\frac{1}{12}\)

Vậy \(x=\frac{1}{12}\)

\(\left(2-3x\right)^2=\frac94\)

\(\left(2-3x\right)^2=\left(\frac32\right)^2\)

\(\Rightarrow2-3x=\frac32\)

\(3x=2-\frac32\)

\(3x=\frac12\)

\(x=\frac12:3\)

\(x=\frac12\cdot\frac13\)

\(x=\frac16\)
Vậy \(x=\frac16\)