a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

bạn ơi sao bài này khó thế?

1+7+7 mũ 2+7 mũ 3......+7 mũ 100.Tính a,a là tổng dãy số trên 

a) 7.x - x = 5²¹ : 5¹⁹ + 3.2² - 7⁰

6x = 25 + 12 - 1

6x = 36

x = 6

b) 7x - 2x = 6¹⁷ : 6¹⁵ + 44 : 11

5x = 36 + 4

5x = 40

x = 8

c) 5x + x = 39 - 3¹¹ : 3⁹

6x = 39 - 9

6x = 30

x = 5

d) [(6x - 39) : 7]. 4 = 12

(6x - 39) : 7 = 12 : 4

(6x - 39) : 7 = 3

6x - 39 = 3 . 7

6x - 39 = 21

6x = 21 + 39

6x = 60

x = 10

x = 

11 mũ 39 > 5 mũ 26 vì cơ số và số mũ đều hơn
2 mũ 125 và 3 mũ 75
2^125=(2^5)^25=32^25
3^75=(3^3)^25=27^25
Vì 32^25>27^25 nên 2^125>3^75
5^40=(5^4)^10=3125^10
Vì 3125^10>620^10 nên 5^40>620^10

b: \(B=\left(1+7\right)+7^2\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(=8\cdot\left(1+7^2+...+7^{100}\right)⋮8\)

c: \(C=4^{39}\left(1+4+4^2\right)=4^{39}\cdot21=4^{38}\cdot84⋮28\)

 Bài 1:

1; 5\(x\) + \(x\) = 39 - 3¹¹ : 3⁹

    \(x\).(5 + 1) = 39 - 3²

    \(x.6\)        = 39 - 9

   \(x.6\)         = 30

   \(x\)           = 30 : 6

   \(x\)          = 5

Vậy \(x\) = 5

2; 5\(x\) + \(x\) = 150 : 2 + 3

    \(x\).(5 + 1) = 75 + 3

    \(x.6\)          = 78

    \(x\)             = 78 : 6

     \(x\)            = 13

Vậy \(x=13\) 

3;      7\(x\) - \(x\) = 5²¹ : 5¹⁹ + 3.2² - 1

    \(x.\left(7-1\right)\) = 5² + 6 - 1

    \(x\).6 = 25 + 6 - 1

    \(x.6\) = 31 - 1

   \(x.6\) = 30

   \(x\)    = 30 : 6

    \(x=5\)

Vậy \(x=5\)

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)

a,\(\dfrac{3^6.5^7.7^{11}}{3^4.5^7.7^{10}}=\dfrac{3^4.3^2.5^7.7^{10}.7}{3^4.5^7.7^{10}}\) \(=9.7=63\)

b,\(\dfrac{2^{43}+2^4}{2^{39}+1}=\dfrac{2^{39}.2^4+2^4}{2^{39}+1}\) \(=\dfrac{2^4\left(2^{39}+1\right)}{2^{39}+1}=16\)