Question

Tìm x, biết:

a)\(\left(\frac{1}{3}\right)^2.27=3^x\)

b)\(\frac{64}{\left(-2\right)^x}=-32\)

c)\(3x^2-\frac{1}{2}x=0\)

 

 

Answer 1

a.

\(\left(\frac{1}{3}\right)^2\times27=3^x\)

\(\frac{1^2}{3^2}\times3^3=3^x\)

\(3^1=3^x\)

\(x=1\)

b.

\(\frac{64}{\left(-2\right)^x}=-32\)

\(\frac{\left(-2\right)^6}{\left(-2\right)^x}=\left(-2\right)^5\)

\(\left(-2\right)^x=\frac{\left(-2\right)^6}{\left(-2\right)^5}\)

\(\left(-2\right)^x=-2\)

\(x=1\)

c.

\(3x^2-\frac{1}{2}x=0\)

\(x\times\left(3x-\frac{1}{2}\right)=0\)

TH1:

\(x=0\)

TH2:

\(3x-\frac{1}{2}=0\)

\(3x=\frac{1}{2}\)

\(x=\frac{1}{2}\div3\)

\(x=\frac{1}{2}\times\frac{1}{3}\)

\(x=\frac{1}{6}\)

Vậy x = 0 hoặc x = 1/6