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\(e,\frac{22}{15}-x=-\frac{8}{27}\)
=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)
=> \(x=\frac{22}{15}+\frac{8}{27}\)
=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)
\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)
=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)
=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)
=> \(\frac{2x-5}{5}=-\frac{5}{4}\)
=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)
=> \(2x=-\frac{5}{4}\)
=> \(x=-\frac{5}{8}\)
\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)
=> \(-\frac{9}{4}x+\frac{37}{4}=20\)
=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)
=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)
\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)
=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)
=> \(-10\le x\le-\frac{13}{7}\)
Đến đây tìm x
a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=9\)
b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)
\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)
\(\Leftrightarrow x=66\)
đúng đó nhưng bạn nên nhóm \(\frac{-22}{3}\)và \(\frac{-5}{3}\)vào với nhau và đặt \(\frac{5}{9}\)ra ngoài sẽ tính nhanh hơn
\(H=\frac{4}{15}-\frac{23}{28}-\left(-\frac{23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}+\frac{-23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}+\frac{-2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}+\frac{-2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=1+\frac{22}{27}\)
\(H=\frac{27}{27}+\frac{22}{27}=\frac{49}{27}\)
\(H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}-\frac{23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}-\frac{2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}-\frac{2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=\frac{49}{27}\)
\(\frac{8^{15}}{4^{22}}.\frac{3^{16}}{9^8}=\frac{\left(2^3\right)^{15}.3^{16}}{\left(2^2\right)^{22}.\left(3^2\right)^8}=\frac{2^{45}.3^{16}}{2^{44}.3^{16}}=\frac{2^{45}}{2^{44}}=2.\)
\(\frac{\left(2^3\right)^{15}^{ }}{\left(2^2\right)^{22}}.\frac{3^{16}}{\left(3^3\right)^8}\)
\(\frac{2^{18}}{2^{44}}.\frac{3^{16}}{3^{24}}\)
e. \(x=\frac{22}{15}-\frac{8}{27}\)
\(x=\frac{158}{135}\)
Vậy \(x\in\left\{\frac{158}{135}\right\}\)
\(\frac{22}{15}-x=\frac{8}{27}\)
\(x=\frac{22}{15}-\frac{8}{27}\)
\(x=\frac{198}{135}-\frac{40}{135}\)
\(x=\frac{158}{135}\)
Vậy \(x=\frac{158}{135}\)
Chúc bạn học tốt !!!