\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{25-3}=\frac{8}{22}=\frac{4}{11}\)

\(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{5-9}{14}=\frac{-4}{14}=\frac{-2}{7}\)

\(\frac{-2}{7}=\frac{-22}{77}\)

\(\frac{4}{11}=\frac{28}{77}\)

= 28/77 

Sửa: \(A=\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)

\(A=\left(3x+2\right)^2-2\left(3x+2\right)\left(2x-7\right)+\left(2x-7\right)^2\)

\(A=\left[\left(3x+2\right)-\left(2x-7\right)\right]^2\)

\(A=\left(3x+2-2x+7\right)^2\)

\(A=\left(x+9\right)^2\)

Thay \(x=-19\) vào A ta có:

\(A=\left(-19+9\right)^2=\left(-10\right)^2=100\)

Vậy: ...

a: \(\dfrac{4\cdot5+4\cdot11}{8\cdot7+4\cdot3}=\dfrac{4\cdot16}{4\left(2\cdot7+3\right)}=\dfrac{16}{17}=\dfrac{11088}{11781}\)

\(\dfrac{-15\cdot8+10\cdot7}{5\cdot6+20\cdot3}=\dfrac{-120+170}{30+60}=\dfrac{50}{90}=\dfrac{5}{9}=\dfrac{6545}{11781}\)

\(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=2\cdot5\cdot\dfrac{1}{77}=\dfrac{10}{77}=\dfrac{1530}{11781}\)