1) 27⁴ x 81¹⁰

= (3³)⁴ x (3⁴)¹⁰

= 3^{3 x 4} x 3^{4 x 10} 

= 3¹² x 3⁴⁰

= 3⁵²

2) a) n¹⁶ = n³

=> n¹⁶ - n³ = 0 

=> n³ x (n¹³ - 1) = 0

=> \(\orbr{\begin{cases}n^3=0\\n^{13}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}n^3=0^3\\n^{13}=1^3\end{cases}\Rightarrow}\orbr{\begin{cases}n=0\\n=1\end{cases}}}\)

Vậy n = 0 hoặc n = 1

\(\left(2x-3\right)^{10x}=\left(3-2x\right)^{100}\)

\(\Leftrightarrow\left(2x-3\right)^{10x}=\left(2x-3\right)^{100}\)

\(\Leftrightarrow10x=100\Leftrightarrow x=100:10=10\)

\(\left(2x-3\right)^{10x}=\left(3-2x\right)^{100}\)

\(\Rightarrow\left(2x-3\right)^{10x}=\left(2x-3\right)^{100}\)

\(\Rightarrow10x\Leftrightarrow x=100:10=10\)

~Study well~ :)

Ta có : 4ⁿ : 2^{n + 3} = 64

=> (2²)ⁿ : 2^{n + 3} = 64

=> 2²ⁿ : 2^{n + 3} = 64

=> 2^{2n - (n + 3)} = 64

=> 2^{2n - n - 3} = 2⁷

=> 2^{n - 3} = 2⁷

=> n - 3 = 7

=> n      = 7 + 3

=> n      = 10

Vậy n = 10 

Theo bài ra ta có : \(\frac{23+n}{40+n}=\frac{3}{4}\)

\(\Rightarrow4.\left(23+n\right)=3.\left(40+n\right)\)

\(\Rightarrow92+4n=120+3n\)

\(\Rightarrow4n-3n=120-92\)

\(\Rightarrow n=28\)

Vậy số n cần tìm là 28

=>( 23+a)/(40+a)=3/4

=>4*(23+a)=3*(40+a)

=> 92+4a=120+3a

=> a=28

=.số N đó là 28

(x + 3)(x² - 36) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x^2-36=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-3\\x=\pm6\end{cases}}\)

=> \(\orbr{\begin{cases}x=-3\\x=\pm6\end{cases}}\)

(x + 3) . (x^2 - 36) = 0

=> TH1: x + 3 = 0

              x       = -3

TH2: x^2 - 36 = 0

         x^2        = 36

         x^2        = 6^2

=> x = 6

Vậy x \(\in\){6 ; -3}

\(1-2b+c-3a=9\)

\(\Leftrightarrow1-2\left(-3\right)+\left(-7\right)-3a=9\)

\(\Leftrightarrow1+6-7-3a=9\)

\(\Leftrightarrow-3a=9\Rightarrow a=\frac{9}{-3}=-3\)

Với b=-3; c=-7 ta được

                1 - 2b + c - 3a = 9

<=>          1 - 2.(-3) + (-7) - 3a = -9

<=>          1 + 6 - 7 - 3a = -9

<=>         -3a = -9

<=>          a  = 3

Ta có: abcd = 100ab + cd

                   = 100.3.cd + cd

                   = 300.cd + cd

                   = (300 + 1).cd

                   = 301.cd

Vì \(301⋮7\Rightarrow abcd⋮7\)

Ta có : abcd = ab00 + cd

                     = 100.cd + cd

                     = 3.100.cd + cd

                      = 300.cd + cd

                       = 301.cd

                       = 7.43.cd \(⋮\)7

\(\Rightarrow\)abcd \(⋮\)7 nếu ab = 3.cd 

Vậy abcd \(⋮\)7 nếu ab = 3.cd (đpcm)

\(D=4+4^2+4^3+4^4+...+4^{200}\)

\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{199}+4^{200}\right)\)

\(=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{199}.\left(1+4\right)\)

\(=\left(1+4\right).\left(4+4^3+...+4^{199}\right)\)

\(=5.\left(4+4^3+...+4^{199}\right)⋮5\)

Lần sau ghi đề hẳn hoi đừng đùa 

cảm ơn bạn

\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)

\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\frac{98}{99}\)

\(\Rightarrow B=\frac{98}{33}\)

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=1-\frac{1}{7}\)

\(\Rightarrow A=\frac{6}{7}\)