đáp án đề số 89
link:http://vanhay.edu.vn/de-thi-thu-thpt-quoc-gia-mon-van-chuan-cau-truc-de-89/
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\(\left(x-3\right)\left(2x+6\right)=0\)
<=> \(\hept{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=-3\end{cases}}\)
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
\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)
\(=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)
\(=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)
\(=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)
\(=\frac{2^{40}}{2^{30}}=2^{10}\)
\(\frac{8^{20}+4^{20}}{4^{25}+64^5}\)
\(=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)
\(=\frac{2^{60}+2^{40}}{2^{25}+2^{30}}\)
\(=\frac{2^{40}\left(2^{20}+1\right)}{2^{25}\left(1+2^5\right)}\)
\(=\frac{2^{15}\left(2^{20}+1\right)}{1+2^5}\)
\(=\frac{2^{35}+2^{15}}{1+2^5}\)
\(\left(\frac{1}{2}\right)^{10}-\left(\frac{1}{4}\right)^{20}\)
\(=\left(\frac{1}{2}\right)^{10}-\left[\left(\frac{1}{2}\right)^2\right]^{20}\)
\(=\left(\frac{1}{2}\right)^{10}-\left(\frac{1}{2}\right)^{40}\)
\(=\left(\frac{1}{2}\right)^{10}-\left(\frac{1}{2}\right)^{10}\cdot\left(\frac{1}{2}\right)^{30}\)
\(=\left(\frac{1}{2}\right)^{10}\cdot\left[1-\left(\frac{1}{2}\right)^{30}\right]\)
\(=\frac{1}{2^{10}}\cdot\frac{1-2^{30}}{2^{30}}\)
\(=\frac{1-2^{30}}{2^{40}}\)
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...
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^6.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^6.2^2.5}=\frac{2^{10}.\left(3^8-3^9\right)}{2^{12}.3^6.5}=\frac{3^8-3^9}{2^2.3^6.5}=\frac{3^7\left(3-3^2\right)}{2^2.3^6.5}=\frac{3.\left(-6\right)}{20}\)\(=\frac{-18}{20}=\frac{-9}{10}\)
\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^6\cdot20}\)
\(=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{12}\cdot3^6\cdot5}\)
\(=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{12}\cdot3^6\cdot5}\)
\(=\frac{3^2\cdot\left(-2\right)}{2^2\cdot5}\)
\(=\frac{-9}{10}\)