5\(x\) + 5⁰ = 3\(x\) + 3²

5\(x\) + 1 = 3\(x\) + 9

5\(x\) - 3\(x\) = 9 -1

2\(x\)        = 8

\(x\)          = 8:2

\(x\)          = 4

5$x$ + 50 = 3$x$ + 32

5$x$ + 1 = 3$x$ + 9

5$x$ - 3$x$ = 9 -1

2$x$        = 8

$x$          = 8:2

$x$          = 4

Ht!

\(5x+5^0=3x+3^2\\ 5x+1=3x+9\\ 5x-3x=9-1\\ x\left(5-3\right)=8\\ x2=8\\ x=8\div2\\ x=4\)

\(a,A=3+3^2+3^3+3^4+...+3^{100}\\ 3A=3^2+3^3+3^4+3^5+3^{101}\\ 3A-A=2A=3^{101}-3\\ \Rightarrow2A+3=3^{101}=3^{4.25+1}\\ \Rightarrow n=25\)

Em kiểm tra lại đề bài, chỗ \(A^2\)

Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$

Bài 2:

a. $7\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 7; -7\right\}$

$\Rightarrow n\in \left\{0; -2; 6; -8\right\}$

b.

$2n+5\vdots n+1$
$\Rightarrow 2(n+1)+3\vdots n+1$

$\Rightarrow 3\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -2; 2; -4\right\}$

\(a,\left(-2+x\right).\left(3x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}\Leftrightarrow}x=2}\)

Vậy ................

\(b,\left(3x+9\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\2x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-3\end{cases}\Leftrightarrow x=-3}\)

Vậy ......................

a)\(\(\left(-2+x\right)\left(3x-6\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}\Rightarrow}x=2}\)\)

b) cmtt

_Minh ngụy_

a,         ( -2 + x ) . ( 3x - 6 ) = 0 

\(\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\3x=6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=2\end{cases}}}\)

b,  ( 3x + 9 ) . ( 2x + 6 ) = 0 

\(\orbr{\begin{cases}3x+9=0\\2x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=-9\\2x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=-3\end{cases}}}\)

Ta có: 2x-10+9-3x<0

=> -x-1<0

<=> -x<1

<=>x>-1