\(A=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^8.\left(2^{10}+1\right)}{1+2^{10}}=2^8\)

\(A=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(A=\frac{2^{12}\cdot\left(2^{18}+2^8\right)}{2^{12}\cdot\left(1+2^{10}\right)}\)

\(A=\frac{2^8\cdot\left(2^{10}+1\right)}{2^{10}+1}\)

Ta có :\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)=\(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)= \(\frac{2^{10}.\left(2^{10}+1\right)}{2^{12}.\left(2^{10}+1\right)}\)=\(\frac{2^{10}}{2^{12}}\)= 4.

phải là 1/4 chứ 

a) \(\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{4}{9}\times2=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{23}{9}\)

b) \(\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{2}{5}\times\dfrac{3}{2}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)

Câu 4 

\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times4\times3\times5\times4\times5}{5\times2\times4\times4\times5\times5}=\dfrac{3\times3}{5\times2}=\dfrac{9}{10}\)

Câu 3:

\(a.\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{15}{9}+\dfrac{8}{9}=\dfrac{23}{9}\)

\(b.\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)

Câu 4:

\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times3\times1}{2\times1\times5}=\dfrac{9}{10}\)

4^4

\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=2^8\)

8^10+4^10/8^4+4^11

=(2^3)^10+(2^2)10/(2^3)^4+(2^2)^11

=2^30+2^20/2^12+2^22

=2^20(2^10+1)/2^12(1+2^10)

=2^8=256

vậy giá trị biểu thức =256

Ta có: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)=\(\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)=\(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)=\(\frac{2^{20}.2^{10}+2^{20}}{2^{12}.2^{10}+2^{12}}\)=\(\frac{2^{20}.\left(2^{10}+1\right)}{2^{12}.\left(2^{10}+1\right)}\)=\(\frac{2^{20}}{2^{12}}=2^8\)

\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)   

\(=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)   

\(=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)   

\(=\frac{2^{30}+2^{20}}{2^{22}+2^{12}}\)   

\(=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\)   

\(=\frac{2^{20}}{2^{12}}\)   

\(=2^8\)   

\(=256\)

`A=sqrt{8+2sqrt7}-sqrt{8-2sqrt7}`

`=sqrt{7+2sqrt7+1}-sqrt{7-2sqrt7+1}`

`=sqrt{(sqrt7+1)^2}-sqrt{(sqrt7-1)^2}`

`=sqrt7+1-sqrt7+1=2`

`B=sqrt{11-6sqrt2}+sqrt{6-4sqrt2}`

`=sqrt{9-2.3.sqrt2+2}+sqrt{4-2.2.sqrt2+2}`

`=sqrt{(3-sqrt2)^2}+sqrt{(2-sqrt2)^2}`

`=3-sqrt2+2-sqrt2=5-2sqrt2`

a) Có x = 2020 => x + 1 = 2021. Thay 2021 = x + 1 vào A

\(A=x^6-\left(x+1\right)^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(A=1\)

b) Có x = -19 => x - 1 = -20 => - ( x - 1 ) = 20. Thay 20 = - ( x - 1) vào B

\(B=x^{10}-\left(x-1\right)x^9-\left(x-1\right)x^8-\left(x-1\right)x^7-...-\left(x-1\right)x^2-\left(x-1\right)x-x+1\)

\(B=x^{10}-x^{10}+x^9-x^9+...+x^2-x^2+x-x+1\)

\(B=1\)

Chúc bạn học tốt!!!