Vì D > C ,  B > A

=> D - A > C - B

=> -1 ( D - A ) < ( C - B ) ( -1 )

=> A - D < B - C

Nhầm A,B,C,D mới đúng

ko biết

cs mk nè fan conan

Mình là fan nè :

Bạn muốn hiểu rõ hơn thì vào link này nhé : https://www.youtube.com/watch?feature=player_embedded&v=xl1ILze5ZpM

\(\left(x-3\right)\left(2x+6\right)=0\)

<=> \(\hept{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\x=-3\end{cases}}\)

học tốt

a. (x - 3) . (2x + 6) = 0

<=> \(\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\2x=-6\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy x = + 3

|4x| - |-13,5| =|-7,5|

=|4x| = 7,5 + 13,5 =21

=>4x=21

=> x=21/4

\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)

\(=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)

\(=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)

\(=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)

\(=\frac{2^{40}}{2^{30}}=2^{10}\)

\(\frac{8^{20}+4^{20}}{4^{25}+64^5}\)

\(=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)

\(=\frac{2^{60}+2^{40}}{2^{25}+2^{30}}\)

\(=\frac{2^{40}\left(2^{20}+1\right)}{2^{25}\left(1+2^5\right)}\)

\(=\frac{2^{15}\left(2^{20}+1\right)}{1+2^5}\)

\(=\frac{2^{35}+2^{15}}{1+2^5}\)

\(\left(\frac{1}{2}\right)^{10}-\left(\frac{1}{4}\right)^{20}\)

\(=\left(\frac{1}{2}\right)^{10}-\left[\left(\frac{1}{2}\right)^2\right]^{20}\)

\(=\left(\frac{1}{2}\right)^{10}-\left(\frac{1}{2}\right)^{40}\)

\(=\left(\frac{1}{2}\right)^{10}-\left(\frac{1}{2}\right)^{10}\cdot\left(\frac{1}{2}\right)^{30}\)

\(=\left(\frac{1}{2}\right)^{10}\cdot\left[1-\left(\frac{1}{2}\right)^{30}\right]\)

\(=\frac{1}{2^{10}}\cdot\frac{1-2^{30}}{2^{30}}\)

\(=\frac{1-2^{30}}{2^{40}}\)

a)  \(\left(2x+3\right)^2=\frac{9}{21}\)

<=>  \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-1\frac{4}{11}\\x=-1\frac{7}{11}\end{cases}}\)

Vậy...

b)  \(\left(3x-1\right)^3=\frac{-8}{27}\)

<=>   \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)

<=>   \(3x-1=\frac{-2}{3}\)

<=>  \(3x=\frac{1}{3}\)

<=>  \(x=\frac{1}{9}\)

Vậy....