Question

3) tìm x:

a) \(x^2=\dfrac{1}{16}\)

b)\(x^5\div x^2=\dfrac{-1}{64}\)

c)\(x^3\div x^2=\dfrac{32}{243}\)

d)\(\left(x^2\right)^2=\dfrac{81}{16}\)

Answer 1

a, \(x^2=\dfrac{1}{16}\Rightarrow x=\pm\dfrac{1}{4}\)

b, \(x^5:x^2=-\dfrac{1}{64}\Rightarrow x^3=\left(-\dfrac{1}{4}\right)^3\Rightarrow x=-\dfrac{1}{4}\)

c, \(x^3:x^2=\dfrac{32}{243}\Rightarrow x=\dfrac{32}{243}\)

d, \(\left(x^2\right)^2=\dfrac{81}{16}\Rightarrow x^4=\left(\dfrac{3}{2}\right)^4\Rightarrow x=\pm\dfrac{3}{2}\)

Chúc bạn học tốt!!!

Answer 2

3) Tìm x

a) \(^{x^2}\)=\(\dfrac{1}{16}\)

<=> x = \(\sqrt{-\dfrac{1}{16}}\)

\(\sqrt{\dfrac{1}{16}}\)

<=> x = -14

+14

b) \(x^{5^{ }}\): \(x^2\) = \(-\dfrac{1}{64}\)

<=> \(^{x^{5-2}}\) =\(-\dfrac{1}{64}\)

<=> \(x^3\) = \(-\dfrac{1}{64}\)

<=> x = \(-\dfrac{1}{4}\)

c)\(x^3:x^2\) = \(\dfrac{32}{243}\)

<=> \(^{x^{3-2}}\) = \(\dfrac{32}{243}\)

<=> x = \(\dfrac{32}{243}\)

d) \((x^2)^2\) = \(\dfrac{81}{16}\)

<=>\(^{x^{2.2}}\) = \(\dfrac{81}{16}\)

<=> \(x^4\) = \(\dfrac{81}{16}\)

<=> x = \(\dfrac{3}{2}\)

\(-\dfrac{3}{2}\)

Answer 3

a) \(x^2=\dfrac{1}{16}\)

=> \(x^2=\pm\left(\dfrac{1}{4}\right)^2\)

=> \(x=\pm\dfrac{1}{4}\)

b) \(x^5:x^2=\dfrac{-1}{64}\)

=> \(x^{5-2}=\dfrac{-1}{64}\)

=> \(x^3=\dfrac{-1}{64}\)

=> \(x^3=\left(\dfrac{-1}{4}\right)^3\)

=> \(x=\dfrac{-1}{4}\)

c) \(x^3:x^2=\dfrac{32}{243}\)

=> \(x^1=\dfrac{32}{243}\)

=> \(x=\dfrac{32}{243}\)

d) \(\left(x^2\right)^2=\dfrac{81}{16}\)

=> \(x^4=\dfrac{81}{16}\)

=> \(x^4=\pm\left(\dfrac{3}{2}\right)^4\)

=> \(x=\pm\dfrac{3}{2}\)