Bài 1:
b) \(\left(3x-\frac{1}{5}\right)^3=\frac{64}{125}\)
=> \(\left(3x-\frac{1}{5}\right)^3=\left(\frac{4}{5}\right)^3\)
=> \(3x-\frac{1}{5}=\frac{4}{5}\)
=> \(3x=\frac{4}{5}+\frac{1}{5}\)
=> \(3x=1\)
=> \(x=1:3\)
=> \(x=\frac{1}{3}\)
Vậy \(x=\frac{1}{3}.\)
c) \(\left(3x-1\right)^3=-\frac{8}{27}\)
=> \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-1=-\frac{2}{3}\)
=> \(3x=\left(-\frac{2}{3}\right)+1\)
=> \(3x=\frac{1}{3}\)
=> \(x=\frac{1}{3}:3\)
=> \(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}.\)
Chúc bạn học tốt!
Bài 1:
a)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{6}\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{1}{6}}\\x+\frac{1}{2}=-\sqrt{\frac{1}{6}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\frac{1}{6}}-\frac{1}{2}\\x=-\sqrt{\frac{1}{6}}-\frac{1}{2}\end{matrix}\right.\)
Vậy...
b) \(\left(3x-\frac{1}{5}\right)^3=\frac{64}{125}\Leftrightarrow\left(3x-\frac{1}{5}\right)^3=\left(\frac{4}{5}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{5}=\frac{4}{5}\Leftrightarrow3x=\frac{5}{5}\Leftrightarrow x=\frac{1}{3}\)
c)\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
Hay \(3x-1=-\frac{2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)
Bài 2:
a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
Do đó \(2^{225}< 3^{150}\)
b) \(3^{34}>3^{30}=\left(3^3\right)^{10}=27^{10}>25^{10}=\left(5^2\right)^{10}=5^{20}\)
Do đó \(3^{34}< 5^{20}\)
c) Chưa nghĩ ra
