\(3^{2x-1}=\frac{1}{729}=\frac{1}{3^6}=3^{-6}\Rightarrow2x-1=-6\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)
\(\text{Vậy:}x=-\frac{5}{2}\)
\(3^{2x-1}=\frac{1}{729}=\frac{1}{3^6}=3^{-6}\Rightarrow2x-1=-6\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)
\(\text{Vậy:}x=-\frac{5}{2}\)
\(\frac{\left(\frac{-1}{9}\right)^0.3^2.9^3}{729}\)
So sánh A=[6.(-1/3)^2-(-1/3)+1]:(-1/3-1) và B=(729-1^3) (729-2^3) (729-3^3)...(729-125^3)
TÌM X:
\(\left(3^x\right)^2:3^2=\frac{1}{729}\)
Viết biểu thức sau dưới dạng lũy thừa rồi tính
\(\left[\left(\frac{1}{9}:\frac{8}{27}\right):\frac{16}{48}\right]:\frac{729}{128}\)
Ai làm được mk tích cho nha
tìm x biết : a,\(\frac{x^7}{81}=27\); b,\(\frac{x^8}{9}=729\); c.\(^{\left[x-\frac{1}{2}\right]^0}\); d,\(\left[X-2\right]^2=1\); e, \(\left[2X-1\right]^3=8\); f,\(\left[X+\frac{1}{2}\right]^2\)=\(\frac{1}{16}\)
1. Tìm x\(\inℚ\), biết:
a)\(\left(x-\frac{1}{2}\right)2=0\)
b)\(\left(x-2\right)^2=1\)
c) \(\left(2x-1\right)^3=-8\)
d)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
e)\(\frac{x^7}{81}=27\)
g)\(\frac{x^8}{9}=^{729}\)
32x-1=27x+2
32x-1=27x+2
32x-1=27x+2
giúp với