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

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{2012}+3^{2013}+3^{2014}+3^{2015}\right)\)

\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{2012}\left(1+3+3^2+3^3\right)\)

\(=\left(1+3+3^2+3^3\right)\left(1+3^4+...+3^{2012}\right)\)

\(=40\left(1+3^4+...+3^{2012}\right)\)\(⋮\)\(5\)

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

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+..+2^{2013}\left(1+2+2^2+2^3\right)\)

\(=\left(1+2+2^2+2^3\right)\left(2+2^5+...+2^{2013}\right)\)

\(=15\left(2+2^5+...+2^{2013}\right)\)\(⋮\)\(15\)

\(a>b\)Vì :\(7256.4375>3650+4375\)

I don't now

mik ko biết 

sorry 

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1)\(4n+3⋮n-2\)

\(\Leftrightarrow4n+3=4\left(n-2\right)+11\)

\(\Rightarrow4\left(n-2\right)⋮n-2\)\(\Rightarrow n-2⋮n-2\)

\(\Rightarrow11⋮n-2\)

\(\Rightarrow n-2\in\left\{\pm1;\pm11\right\}\)

\(\Rightarrow n\in\left\{3;1;13;-9\right\}\)

2)\(xy+5x+y+10=0\)

\(\Leftrightarrow x\left(y+5\right)+y+5+5=0\)

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

\(\Leftrightarrow\left(x+1\right).\left(y+5\right)=-5\)

3)

I don't now

mik ko biết 

sorry 

......................

I don't now

mik ko biết 

sorry 

......................

ĐKXĐ:  \(x\ne0;x\ne-2\)

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)

\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)

\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}=\frac{1}{18}\)

\(\Rightarrow\)\(x+2=18\)

\(\Leftrightarrow\)\(x=16\) (t/m ĐKXĐ)

Vậy...

1/2(1-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2)=4/9

1/2(1-1/x+2)=4/9

1-  1/x+2=4/9:1/2

1  -  1  /x+2=8/9

1/x+2=1-8/9

1/x+2=1/9

suy ra x+2=9

        x=9-2

        x=7

a)\(51-\left|x-1\right|=-\left(-44\right)=44\)

\(\Rightarrow\left|x-1\right|=51-44=7\)

\(\Rightarrow\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)

b)\(6x-5\left(x-7\right)=\left(27-51\right).\left(486-73\right)\)

\(\Leftrightarrow6x-\left(5x-35\right)=-9912\)

\(\Leftrightarrow6x-5x+35=-9912\)

\(\Rightarrow x+35=-9912\)

\(\Rightarrow x=-9912-35=-9947\)

I don't now

mik ko biết 

sorry 

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