\(A=\frac{2^{23}-2^{21}\cdot2^2\cdot5}{2^{24}}\)

\(A=\frac{2^{23}-2^{23}\cdot5}{2^{24}}\)

\(A=\frac{2^{23}\left(1-5\right)}{2^{24}}\)

\(A=\frac{-\left(2^{23}\cdot2^2\right)}{2^{24}}=-\frac{2^{25}}{2^{24}}=-2\)

A =5.2⁴ - (3² + 1)²¹ : 10²⁰

A = 5.16 - (9 + 1)²¹: 10²⁰

A = 5.16 - 10²¹ : 10²⁰

A = 5.16 - 10

A = 80 - 10

A = 70

\(A=\frac{1}{2}+\frac{-1}{7}-\frac{-1}{13}+\frac{-1}{13}-\frac{-2}{5}+\frac{-11}{21}+\frac{1}{10}\)

\(A=\frac{5}{10}-\frac{3}{21}+\frac{1}{13}-\frac{1}{13}+\frac{4}{10}-\frac{11}{21}+\frac{1}{10}\)

\(A=\left(\frac{5+4+1}{10}\right)+\left(\frac{-3}{21}-\frac{11}{21}\right)+\left(\frac{1}{13}-\frac{1}{13}\right)\)

\(A=1+\frac{-2}{3}=\frac{3-2}{3}=\frac{1}{3}\)

\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=3\)

\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=\frac{2^2.78}{104}=\frac{2^2.2.39}{2^3.13}=\frac{2^3.39}{2^3.13}=3\)

\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(=>A=\frac{2^{10}.\left(65+35\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=3\)

\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

  \(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{6}{21}\)

tui chả hiểu bạn nói gì cả

\(A=\frac{10-1\frac{1}{6}\times\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{11}}\)

\(A=\frac{10-\frac{7}{6}\times\frac{6}{7}}{21:\frac{11}{2}+\frac{57}{11}}\)

\(A=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}\)

\(A=\frac{9}{9}=1\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)

\(=\frac{2^{10}.78}{2^8.104}\)

\(=\frac{2^8.2^2.78}{2^8.104}\)

\(=\frac{2^8.4.78}{2^8.104}\)

\(=\frac{2^8.312}{2^8.104}\)

\(=\frac{2^8.3.104}{2^8.104}=3\)

21. Phân tích A thành \(A=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(a^2+b^2+c^2+ab+bc+ac\right)\). Từ đó dễ dàng chứng minh.

23. \(9y\left(y-x\right)=4x^2\Leftrightarrow9y^2-9xy=4x^2\Leftrightarrow4x^2+9xy-9y^2=0\)

Chia cả hai vế của đẳng thức trên với \(y^2>0\)được : 

\(4\left(\frac{x}{y}\right)^2+\frac{9x}{y}-9=0\). Đặt \(t=\frac{x}{y},t>0\)(Vì x,y dương)

\(\Rightarrow4^2+9t-9=0\Leftrightarrow\orbr{\begin{cases}t=\frac{3}{4}\left(\text{nhận}\right)\\t=-3\left(\text{loại}\right)\end{cases}}\)

Vậy \(\frac{x}{y}=\frac{3}{4}\Rightarrow y=\frac{4x}{3}\)thay vào biểu thức được :

\(\frac{x-y}{x+y}=\frac{x-\left(\frac{4x}{3}\right)}{x+\left(\frac{4x}{3}\right)}=-\frac{1}{7}\)

24. Tương tự câu 23 , ta được \(x=y\) hoặc \(y=-3x\)(loại trường hơp này vì mẫu thức phải khác 0)

 Vậy với x = y được \(A=-\frac{1}{2}\)

\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)

\(=>\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)

\(=>1+\frac{16}{21}-\frac{20}{15}\)

\(=>\frac{37}{21}-\frac{20}{15}\)

\(=>\frac{3}{7}\)

\(b,12-8\cdot\left(\frac{3}{2}\right)^3\)

\(=>12-8\cdot\frac{27}{8}\)

\(=>12-27\)

\(=>-15\)

\(c,\left(\frac{1}{9}\right)^{2005}\cdot9^{2005}-96^2:24^2\)

\(=>\left(\frac{1^{2005}^{ }}{9^{2005}}\cdot9^{2005}\right)-\left(96^2:24^2\right)\)

\(=>\left(1^{2005}\right)-16\)

\(=>1-16\)

\(=>-15\)