Question

Tìm x:

a)\(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\)

b)\(81^{-2x}.27^x=9^5\)

c)\(2^x+2^{x+3}=288\)

Answer 1

a.\(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\)

\(\Rightarrow\left(-\frac{1}{6}\right)^{x-1}=\frac{1}{36}\)

\(\Rightarrow \left(-\frac{1}{6}\right)^{x-1}=\left(-\frac{1}{6}\right)^2\)

=> x-1=2

=> x=2+1

Vậy x=3.

b.\(81^{-2x}.27^x=9^5\)

\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=\left(3^2\right)^5\)

\(\Rightarrow3^{4.\left(-2x\right)}.3^{3x}=3^{10}\)

\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)

\(\Rightarrow3^{-5x}=3^{10}\)

=> -5x=10

=> x=10:(-5)

Vậy x=-2.

c.\(2^x+2^{x+3}=288\)

\(\Rightarrow2^x.\left(1+2^3\right)=288\)

\(\Rightarrow2^x.9=288\)

\(\Rightarrow2^x=288:9\)

\(\Rightarrow2^x=32\)

=> 2^x=2⁵

Vậy x=5.

Answer 2

1234567890357159951753,./asd