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\(\dfrac{81}{125}-\dfrac{8}{27}=\dfrac{3^4}{5^3}-\dfrac{2^3}{3^3}\)
\(\dfrac{81}{125}=\dfrac{3^4}{5^3}\) ; \(-\dfrac{8}{27}=-\left(\dfrac{2}{3}\right)^3\)
\(\dfrac{81}{125}=\dfrac{3^4}{5^3}\)
\(\dfrac{-8}{27}=\dfrac{\left(-2\right)^3}{3^3}=\left(\dfrac{-2}{3}\right)^3\)
Ta có:
\(\dfrac{81}{125}=\dfrac{3^4}{5^4}=\left(\dfrac{3}{5}\right)^4\)
\(-\dfrac{8}{27}=-\dfrac{2^3}{3^3}=\left(-\dfrac{2}{3}\right)^3\)
d)\(-\frac{8}{27}=\frac{\left(-2\right)^3}{3^3}=\left(-\frac{2}{3}\right)^3\)
h)\(-\frac{27}{64}=\frac{\left(-3\right)^3}{4^3}=\left(-\frac{3}{4}\right)^3\)
\(\frac{-8}{27}=\frac{\left(-2\right)^3}{3^3}\)
\(\frac{-27}{64}=\frac{-3^3}{4^3}\)
Bài 3:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=\pm1\)
+) \(x-2=1\Rightarrow x=3\)
+) \(x-2=-1\Rightarrow x=1\)
Vậy \(x=3\) hoặc \(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vạy \(x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=\frac{-1}{4}\)
Vậy \(x=\frac{-1}{4}\)
Cách viết khác : \(\frac{16}{81}=\frac{2^4}{3^4}=\left(\frac{2}{3}\right)^4\)
Chúc bạn học tốt ^^
\(2^6\)\(0,5^2\)\(\left(\frac{1}{2}\right)^4\)\(\left(\frac{1}{2}\right)^8\)\(\left(\frac{11}{12}\right)^2\)
\(-\frac{8}{27}=-\left(\frac{2}{3}\right)^3.\)
-8/27 = (-2/3)3 nha còn cái kia mình cụng ko bít đâu mình cũng đang hỏi