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a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(\frac{x}{\left(-\frac{1}{3}\right)^3}=-\frac{1}{3}\Rightarrow x=\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^4\)
\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)
=> \(x=\frac{\left(\frac{4}{5}\right)^7}{\left(\frac{4}{5}\right)^5}=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}=\left(\pm\frac{1}{4}\right)^2\)
=> \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
(3x + 1)3 = -27 => (3x + 1)3 = (-3)3 => 3x + 1 = -3 => 3x = -4 => x = -4/3
a)\(x:\left(\frac{-1}{3}\right)^3=\frac{-1}{3}\)
\(=>x:\frac{-1}{27}=\frac{-1}{3}\)
\(=>x=\frac{-1}{3}.\frac{-1}{27}=>x=\frac{1}{81}\)
b) \(\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)
\(=>x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=>x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(=>\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\\\left(x+\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)^2\end{cases}}\)
\(=>\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=\frac{-1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-1\end{cases}}}\)
d|) \(\left(3x+1\right)^3=-27\)
\(=>\left(3x+1\right)^3=\left(-3\right)^3\)
\(=>3x+1=-3\)
\(=>3x=-4=>x=\frac{-4}{3}\)
cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt:>
\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)
bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a)
\(x=\left(\frac{3}{7}\right)^7:\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
b)
\(-\frac{1}{27}\cdot x=\frac{1}{81}\)
\(x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(x=-\frac{1}{3}\)
c)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
d)
\(\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{3}\\x+\frac{1}{2}=\frac{-2}{3}\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\\x=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\end{cases}}\)
Sửa lại câu a : ( nhìn sai số )
\(\frac{3^5}{5^5}\cdot x=\frac{3^7}{7^7}\)
\(x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(x=\frac{3^7}{7^7}\cdot\frac{5^5}{3^5}\)
\(x=\frac{5^5\cdot3^2}{7^7}\)
\(x=\frac{28125}{823453}\)