\(\left(5+5^2+5^3+...+5^{10}\right)+4x-1=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)

\(\Leftrightarrow\left(1+5+5^2+5^3+...+5^{10}\right)+4x-2=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)(1)

1./ Trước tiên, ta tính: 

\(S=1+5+5^2+5^3+...+5^{10}\)

\(\left(5-1\right)\cdot S=\left(5-1\right)\left(1+5+5^2+5^3+...+5^{10}\right)\)

\(\Leftrightarrow4S=5^{11}-5^{10}+5^{10}-5^9+...+5-1=5^{11}-1\)

\(\Leftrightarrow S=\frac{5^{11}-1}{4}=\frac{1}{4}5^{11}-\frac{1}{4}\)

2./ (1) trở thành:

\(\Leftrightarrow\frac{1}{4}5^{11}-\frac{1}{4}+4x-2=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)

\(\Leftrightarrow4x-\frac{1}{2}x=5+\frac{1}{4}\Leftrightarrow\frac{7}{2}x=\frac{21}{4}\)

\(\Leftrightarrow x=\frac{3}{2}\).

4. (x-3) - x = 3

4x - 12 - x = 3

3x - 12 = 3

3x = 3+12

3x = 15

x= 15:3

x = 5

=56/3

a) A=2¹+2²+2³+...+2²⁰¹⁰

    A=(2¹+2²)+(2³+2⁴)+.....+(2²⁰⁰⁹+2²⁰¹⁰)

    A=(2¹x3)+(2³x3)+.....+(2²⁰⁰⁹x3)

    A=3x(2¹+2³+.......+2²⁰⁰⁹)

Vậy A chia hết cho 3.

NHỮNG CÂU CÒN LẠI BẠN LÀM TƯƠNG TỰ !

A = (3^5.1 + 3^5.3) + (3^7.1 + 3^7.3) + (3^9.1 + 3^9.3)

   = 3^5.(1+3) +  3^7.(1+3) + 3^9.(1+3)

   = 3^5.4 + 3^7.4 + 3^9.4

   =4.(3^5 +3^7 +3^9)

   =4.(3^5.1 +3^5.3^2 + 3^5.3^4)

   =4.3^5.(1+3^2+3^4)

   =4.243.(1+9+81)

   =972.91 chia hết cho 91

Vậy A chia hết cho 91

M=3^5*(1+3+3^2+3^3+3^4+3^5)

M=3^5*[(1+3^2+3^4)+(3+3^3+3^5)]

M=3^5*[91+273]

M=3^5*91*4

=>M chia het cho 91

TH1 : \(91-3x< 7+x\Rightarrow3x+x>91-7\Rightarrow4x>84\Rightarrow x>21\left(1\right)\)

TH2 : \(7+x\ge64\Rightarrow x\ge57\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow x\ge57\)

91 - 3\(x\) < 7 +  \(x\) ≥ 64

⇒ \(\left\{{}\begin{matrix}91-3x< 7+x\\7+x\ge64\end{matrix}\right.\) 

     \(\left\{{}\begin{matrix}7+x+3x>91\\x\ge64-7\end{matrix}\right.\)

      \(\left\{{}\begin{matrix}4x>91-7\\x\ge64-7\end{matrix}\right.\)

       \(\left\{{}\begin{matrix}4x>84\\x\ge57\end{matrix}\right.\)

        \(\left\{{}\begin{matrix}x>84:4\\x\ge57\end{matrix}\right.\)

         \(\left\{{}\begin{matrix}x>21\\x\ge57\end{matrix}\right.\)

          \(x\ge\) 57