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19 tháng 10 2022

\(A=\left(9.25.32\right):\left(8.9\right)\\ =7200:72\\ =100\)

20 tháng 10 2022

\(A=\left(3^2.5^2.2^5\right):\left(2^3.3^2\right)\\ =\left(9.25.32\right):\left(8.9\right)\\ =7200:72\\ =1000\)

Bài 3:

a: a*S=a^2+a^3+...+a^2023

=>(a-1)*S=a^2023-a

=>\(S=\dfrac{a^{2023}-a}{a-1}\)

b: a*B=a^2-a^3+...-a^2023

=>(a+1)B=a-a^2023

=>\(B=\dfrac{a-a^{2023}}{a+1}\)

6 tháng 5 2022

a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)

b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)

c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)

6 tháng 5 2022

d)\(\dfrac{a}{b}=3\times\dfrac{2}{7}=\dfrac{6}{7}\)

e)\(\dfrac{a}{b}=\dfrac{7}{5}-\left(\dfrac{2}{5}\times\dfrac{1}{2}\right)=\dfrac{7}{5}-\dfrac{1}{5}=\dfrac{6}{5}\)

15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9

Sửa đề: 

A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007

=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007

=-1+1+1/2007=1/2007

16:

\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)

=1/64

17:

=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7

=-6/7

a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=4-2\sqrt{2}\)

b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)

 

24 tháng 8 2023

Sai rồi đấy ạ  câu P

12 tháng 10 2023

\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)

\(A=2^{21}-2\)

B tương tự câu A

\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)

\(C=\dfrac{5^{51}-5}{4}\)

\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)

\(D=\dfrac{3^{101}-1}{2}\)

12 tháng 10 2023

\(A=2^1+2^2+2^3+...+2^{20}\)

\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)

\(A=2^{21}-2\)

 

\(B=2^1+2^3+2^5+...+2^{99}\)

\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)

\(B=\)\(\left(2^{101}-2\right):3\)

 

\(C=5^1+5^2+5^3+...+5^{50}\)

\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)

\(C=(5^{51}-5):4\)

 

\(D=3^0+3^1+3^2+...+3^{100}\)

\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)

\(D=(3^{101}-1):2\)

4 tháng 11 2023

Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

\(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

\(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

\(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

\(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

\(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

\(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

\(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

\(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

\(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

\(\dfrac{29}{6}\)

4 tháng 11 2023

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

\(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

\(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

\(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

\(\dfrac{43}{34}\)

  

6 tháng 8 2016

1)

a. \(\left(3x^2-50\right)^2=5^4\)

\(\Leftrightarrow3x^4-50=625\)

\(\Leftrightarrow3x^4=675\)

\(\Leftrightarrow x^4=225\)

\(\Leftrightarrow x=\sqrt{15}\) 

2)

a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)

\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)

b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)

\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)