Hai bài trên áp dụng công thức với khoảng cách là 2.

Ta có:

\(D=1+2^1+2^2+2^3+.....+2^{150}\)

\(\Rightarrow2D-D=\left(2+2^2+2^3+2^4+.....+2^{151}\right)-\left(1+2+2^2+2^3+....+2^{150}\right)\)

\(\Rightarrow D=2^{151}-1\)

\(E=1+4^1+4^2+....+4^{400}\)

\(\Rightarrow4E-E=\left(4+4^2+4^3+....+4^{401}\right)-\left(1+4^1+4^2+....+4^{400}\right)\)

\(\Rightarrow E\left(4-1\right)=4^{401}-1\Leftrightarrow E=\frac{4^{401}-1}{4-1}\)

Các câu còn lại làm tương tự

a) \(\left(3^4.57-9^2.21\right):3^5\)

\(=\left(3^4.57-3^4.21\right):3^5\)

\(=\left[3^4\left(57-21\right)\right]:3^5\)

\(=3^4.36:3^5\)

\(=3^4.2^2.3^2:3^5\)

\(=3.4\)

\(=12\)

b) Ta có; \(1^3+2^3+...+9^3=2025\)

\(\Leftrightarrow2^3.\left(1^3+2^3+....+9^3\right)=2^3.2025\)

\(\Leftrightarrow2^3+4^5+...+18^3=16200\)

Ta có : \(2^3+4^3+6^3+...+18^3\)

\(=2^3\left(1^3+2^3+...+9^3\right)\)

\(=8.2025\)

\(=16200\)

Vậy tổng trên bằng 16200

Tran Phuong Chi            

đáp án là 16200

Bài 8:

a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)

b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Bài 3

2\(^x\) = 16

2\(^x\) = 2⁴

  \(x=4\)

Vậy \(x=4\)

A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 

2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100

2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - (  2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 ) 

A = 2^100 - 2^3 

B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 

5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 

5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )

4B = 5^51 - 1 

B = 5^51 - 1 / 4

a, (-0,2)² \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)

= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)

= 0,2 - \(\dfrac{2}{3}\)

= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)

=  - \(\dfrac{7}{15}\)

b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)

 = \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)

= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)

= \(\dfrac{1}{2}\)

a, (-0,2)2 $\times$ 5 - 213×27346×9546×95213×273​

= 0,04 $\times$ 5 - 213×39212×310212×310213×39​

= 0,2 - 2332​

= 210102​ - 2332​

=  - 715157​

b, 56+22.253+23.125226.5626.5656+22.253+23.1252​

 = 56+4.56+8.5626.5626.5656+4.56+8.56​

= 56.(1+4+8)26.5626.5656.(1+4+8)​

= 1221​
 

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 \(x\in\left\{4;5;6\right\}\)

\(\left(x+1\right)^3=27\)

\(\left(x+1\right)^3=3^3\)

\(\Rightarrow x+1=3\)

\(x=2\)

\(\left(x+1\right)^3=27\)

\(< =>\left(x+1\right)^3=3.3.3=3^3\)

\(< =>x+1=3< =>x=3-1=2\)

\(\left(2x+3\right)^3=9.81\)

\(< =>\left(2x+3\right)^3=9.9.9\)

\(< =>\left(2x+3\right)^3=9^3\)

\(< =>2x+3=9< =>2x=6\)

\(< =>x=\frac{6}{2}=3\)

giúp mình bài này đi

21.73 + 27 . 21

1) 6² . x = 36

    =>  x = 36 / 36 = 1

2) 9x - 4² = 11

    9x - 16 = 11

    9x        = 11 + 16 =27

=>  x        = 27/9=3

3) 10 + 2x = 4²

            2x = 16 - 10 =6

=>          x = 3

4) 231 - ( x - 6 ) =103

             ( x - 6 ) = 231-103=128

           => x = 128+6=134

5) 10 + 3x = 4⁵ / 4²

            3x = 4³-10=64-10=54

              x  = 18

6)25 + 5²x = 8² + 6²

  25  + 25x = 100

           25x = 100-25=75

    =>       x = 75/25=3

*Chúc bạn học tốt*

1) 6².x=36                 2) 9.x-4²=11                3)10+2.x=4²                   4)231-(x-6)=103              5)10+3x=4⁵:42         6)25+5².x=8²+6²

    36.x=36                     9.x-16=11                   10+2.x=16                           x-6 = 231-103          10+3x=4³                   25+25.x =64+36

         x=36:36                9.x     =11+16                   2.x =16-10                     x-6 =128                   10+3x=64               25+25.x =100

         x=1                       9.x     =27                         2.x =6                            x     =128+6                     3x=64-10                25.x =100-25

                                         x     =27:9                         x =6:2                         x      =134                         3x=54                     25.x =75

                                         x     =3                              x =3                                                                       x=54:3                       x =75:25

                                                                                                                                                                x=18                          x =3

zxcvbnm,./