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

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

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

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....

\(\frac{4^5.9^4-2.6^9}{2^{10}.3^6.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^6.2^2.5}=\frac{2^{10}.\left(3^8-3^9\right)}{2^{12}.3^6.5}=\frac{3^8-3^9}{2^2.3^6.5}=\frac{3^7\left(3-3^2\right)}{2^2.3^6.5}=\frac{3.\left(-6\right)}{20}\)\(=\frac{-18}{20}=\frac{-9}{10}\)

\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^6\cdot20}\)

\(=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{12}\cdot3^6\cdot5}\)

\(=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{12}\cdot3^6\cdot5}\)

\(=\frac{3^2\cdot\left(-2\right)}{2^2\cdot5}\)

\(=\frac{-9}{10}\)

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

\(\Rightarrow x^8=x^7\)

\(\Rightarrow x^8-x^7=0\)

\(\Rightarrow x^7\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^7=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vì :\(x\ne0\Rightarrow x=1\)

b)\(x^{10}=25.x^8\)

\(\Leftrightarrow x^{10}-5^2.x^8=0\)

\(\Rightarrow x^8\left(x^2-5^2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^8=0\\x^2-5^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=5^2=25\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Từ I hạ IG; IK lần lượt vuông góc với AC; AB

Do BI; CI là phân giác góc và C nên IH=IG=IK

=> HC=GC=3 (cm) ; HB=KB=2 (cm)

Dễ dàng chứng minh 2 tam giác AKI và AGI là 2 tam giác vuông cân

=> IG=AG; IK=AK. Mà IH=IK=IG => AG=AK=IH=1 (cm)

=> CABC= AK+KB+HB+HC+AG+GC=1+2+2+3+1+3=12 (cm).

thanks