\(2\times99+2\)

\(=2\times\left(99+1\right)\)

\(=2\times100\)

\(=200\)

Ta có:\(C=1+2+2^2+2^3+...+2^{11}\)

\(=>2C=2+2^2+2^3+2^4+...+2^{12}\)

\(=>2C-C=2^{12}-1\)

\(=>C=2^{12}-1\)

Mà \(2^{12}-1\) lẻ \(=>\) C lẻ

\(=>\) C không chia hết cho 2,18

Mặt khác:

\(C=2^{12}-1=\left(2^3\right)^4-1=8^4-1\equiv\left(-1\right)^4-1\equiv0\left(mod9\right)\)

Vậy C chia hết cho 9 (đpcm)

a) \(261\times m+3105\times n\)

Ta có:

\(261⋮9\Rightarrow261\times m⋮9\)

\(3105⋮9\Rightarrow3105\times n⋮9\)

\(\Rightarrow261\times m+3105\times n⋮9\)

Mà \(261\times m+3105\times n>9\)

Vậy \(261\times m+3105\times n\) là bội của \(9\)

b) \(123\times m+1101\times n\)

Ta có:

\(123⋮3\Rightarrow123\times m⋮3\)

\(1101⋮3\Rightarrow1101\times n⋮3\)

\(\Rightarrow123\times m+1101\times n⋮3\)

Mà \(123\times m+1101\times m>3\)

Vậy...(kết luận tương tự câu a nhé)

\(2^{x+2}-2^x=96\)

\(\Rightarrow2^x.2^2-2^x=96\)

\(\Rightarrow2^x\left(2^2-1\right)=96\)

\(\Rightarrow2^x.3=96\)

\(\Rightarrow2^x=96:3\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

\(5^n+5^{n+1}=30\)

\(\Rightarrow5^n+5^n+5=30\)

\(\Rightarrow5^n+5^n=30-5\)

\(\Rightarrow5^n+5^n=25\)

\(\Rightarrow5^n+5^n=5^2\)

\(\Rightarrow n+n=2\)

\(\Rightarrow2n=2\)

\(\Rightarrow n=2:2\)

\(\Rightarrow n=1\)

\(5^{2x}.5^x=625\)

\(\Rightarrow5^{2x}.5^x=5^4\)

\(\Rightarrow2x.x=4\)

\(\Rightarrow2x^2=2^2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=2:2\)

\(\Rightarrow x=1\)

\(2\left(x-3\right)+5\left(x+4\right)=49\)

\(\Rightarrow2x-6+5x+20=49\)

\(\Rightarrow2x+5x=49+6-20\)

\(\Rightarrow7x=35\)

\(\Rightarrow x=35:7\)

\(\Rightarrow x=5\)

plays

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

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

\(=2x^2\left(x^3-x^2+x-1\right)-\left(x^2-x+1\right)\left[x^2-\left(x^2-x-1\right)^2\right]\)

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

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

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

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

e) \(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x+1\right)\)

f) \(xz+yz-5\left(x+y\right)\)

\(=z\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(z-5\right)\)