Question

tìm x biết 

(3x-2)\(^5\) =-243

\(\frac{1}{9}.27^x=3^x\)

\(\frac{1}{3^2}.3^4.3^n=3^7\)

Answer 1

a) ...

(3x-2)^5=(-3)^5

=) 3x-2=(-3)

  3x=(-1)

   x=(-1/3)

DUYỆT CHO MÌH ĐI RỒI MÌH LẠI GIẢI TIẾP CHO

 

Answer 2

\(\left(3x-2\right)^5=-243\)

=> \(\left(3x-2\right)^5=\left(-3\right)^5\)

=> 3x - 2 = -3

=> 3x = -3 + 2

=> 3x = -1

=> x = -1/3

\(\frac{1}{9}.27^x=3^x\)

=> \(3^{-2}.\left(3^3\right)^x=3^x\)

=> 3^-2.3^3x=3^x

=> 3^3x-2=3^x

=> 3x - 2 = x

=> 3x - x = 2

=> 2x = 2

=> x = 1

\(\frac{1}{3^2}.3^4.3^n=3^7\)

=> \(3^{-2}.3^4.3^n=3^7\)

=> 3^n+4-2=3⁷

=> n + 4 - 2 = 7

=> n = 7 + 2 - 4

=> n = 5

Answer 3

1.(3x-2)⁵=-243=(-3)⁵

=>3x-2=-3

=>3x=-3+2=-1

=>x=-1/3

2.1/9.27^x=3^x

=>\(\frac{1}{9}=3^x:27^x=\frac{3^x}{27^x}=\left(\frac{3}{27}\right)^x=\left(\frac{1}{9}\right)^x\Rightarrow x=1\)

3.1/3².3⁴.3ⁿ=3⁷

=>3^-2.3⁴.3ⁿ=3⁷

=>3^-2+4+n=3⁷

=>(-2)+4+n=7

=>2+n=7=>n=5

Vậy n=5