a,\(5^3.2-100:4+2^3.5\)

= 125 . 2 - 25 + 8 . 5

= 250 - 25 + 40

= 265

b, \(6^2:9+50.2-3^3.3\)

= 36 : 9 + 100 - 27 . 3

= 4 + 100 - 81

= 23

các bạn giải bài nhanh cho mình , mình sắp thi học kì 1 rồi

may cai bai day ma cung khong biet oc cho

a) 89-(73-x)=20

=>73-x=89-20

=>73-x=69

=>x=73-69

=>x=4

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.

1. 5³ = 5.5.5 = 125

2. 2⁷ = 2.2.2.2.2.2.2 = 128

3. 4⁴ = 4.4.4.4 = 256

4. 7³ = 7.7.7 = 343

6. 3⁵ = 243

7. 2⁶ =  64

8. 3⁴ =  81

9. 8³ =  512

11. 13² = 169

12. 11² = 121

13. 14² = 196

14. 15² = 225

16. 17² = 289

17. 18² = 324

18. 19² = 361

19. 20² = 400

21. 10⁴ = 10000

22. 10⁵ = 100000

23. 10⁶ = 1000000

24. 10⁷ = 10000000

bạn làm như bạn gia hân là đúng nhé

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

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

a,\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)

\(=\left(\dfrac{13}{14}\right)^2\)

\(=\dfrac{169}{196}\)

b,\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=\left(\dfrac{-1}{12}\right)^2\)

\(=\dfrac{1}{144}\)

c,\(\dfrac{5^4.20^4}{25^5.4^5}\)

\(=\dfrac{100^4}{100^5}\)

\(=\dfrac{1}{100}\)

d,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)

\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)

\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)

\(=4^4.\left(\dfrac{-10}{3}\right)\)

\(=256.\left(\dfrac{-10}{3}\right)\)

\(=\dfrac{-2560}{3}\)

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.\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)

=\(\left(\frac{6}{14}+\frac{7}{14}\right)^2\)

=\(\left(\frac{13}{14}\right)^2\)

=\(\frac{13^2}{14^2}\)

=\(\frac{169}{196}\)

b.\(\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

=\(\left(\frac{9}{12}-\frac{10}{12}\right)^2\)

=\(\left(\frac{-1}{12}\right)^2\)

=\(\frac{-1^2}{12^2}\)

=\(\frac{1}{144}\).

c.Phần C bn viết lại đề bài đi,mk ko hiểu

d.\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)

=\(\frac{-10^5}{3^5}.\left(\frac{-6^4}{5^4}\right)\)

=\(\frac{-100000}{243}.\frac{1296}{625}\)

=\(\frac{-2560}{3}\)

                 Không biết đúng ko nữa

c.\(\frac{5^4.20^4}{25^5.4^5}\)

=\(\frac{25^2.\left(5.4\right)^4}{25^5.4^5}\)

=\(\frac{1.\left(5\right)^4.\left(4\right)^4}{25^3.4^5}\)

=\(\frac{25^2.1}{25^3.4}\)

=\(\frac{1}{25.4}\)

=\(\frac{1}{100}\).