a) \(\dfrac{x^7}{81}=27\) => \(x^7=81.27=3^4.3^3=3^7\)=> \(x=3\)

b) \(\dfrac{x^8}{9}=729\)=> \(x^8=9.729=\)(\(\pm\)\(3^2\)).(\(\pm\)\(3^6\))=(\(\pm\)\(3^{^8}\)) => x = \(\pm\)3

a) \(\frac{x^7}{81}=27\)

=> x⁷ = 27.81

=> x⁷ = 3³.3⁴

=> x⁷ = 3⁷

=> x = 3

Vậy x = 3

b) \(\frac{x^8}{9}=729\)

=> x⁸ = 729.9

=> x⁸ = 3⁶.3²

=> x⁸ = 3⁸ = (-3)⁸

=> \(x\in\left\{3;-3\right\}\)

Vậy \(x\in\left\{3;-3\right\}\)

\(\frac{x^7}{81}=27\Rightarrow x^7=27.81\Rightarrow x^7=3^3.3^4\Rightarrow x^7=3^7\Rightarrow x=3\)

\(\frac{x^8}{9}=729\Rightarrow x^8=729.9\Rightarrow x^8=3^6.3^2\Rightarrow x^8=3^8\Rightarrow x=3\)

x^7/81 = 27                                                    x^8/9 = 729

x^7 : 81= 27                                                   x^8  : 9 = 729

x^7       =27.81                                                x^8       = 729.9

x^7        =2187                                                x^8       =6561

x^7        = 3^7                                                  x^8       = 3^8

=> x      =3                                                       => x      = 3

\(\frac{x^7}{81}=27\Rightarrow x^7=27.81=2187\Rightarrow x=\sqrt[7]{2187}=3\)

\(\frac{x^8}{9}=729\Rightarrow x^8=729.9=6561\Rightarrow x=\sqrt[8]{6561}=3\)

\(\frac{x^7}{81}=27\)

\(\Rightarrow x^7.1=81.27\)

\(\Rightarrow x^7=2187\)

Ta thấy : \(3^7=2187\)

\(\Rightarrow x=\sqrt[7]{2187}=3\)

\(\frac{x^8}{9}=729\)

\(\Rightarrow x^8.1=729.9\)

\(\Rightarrow x^8=6561\)

\(\Rightarrow x=\sqrt[8]{6561}=3\)

a) Ta có: \(\frac{x^7}{81}=27\)

\(\Rightarrow x^7=27.81=2187\)

Mà \(2187=3^7\) \(\Rightarrow x^7=3^7\Leftrightarrow x=3\)

Vậy x = 3

b) Ta có: \(\frac{x^8}{9}=729\)

\(\Rightarrow x^8=729.9=6561\)

Mà \(6561=3^8\) \(\Rightarrow x^8=3^8\Leftrightarrow x=3\)

Vậy x = 3

CHÚC BẠN HỌC TỐT

\(\frac{x^7}{81}=27\Rightarrow27.81=x^7\Rightarrow x=3\)

\(\frac{x^8}{9}=729\Rightarrow x^8=729.9\Rightarrow x=3\)

tíc mình

nha

\(a.\)\(\frac{x^7}{81}=27\)

\(\Rightarrow x^7=27.81\)

\(\Leftrightarrow x^7=2187\)

\(\Leftrightarrow x^7=3^7\)

\(\Leftrightarrow x=3\)

\(b.\)\(\frac{x^8}{9}=729\)

\(\Rightarrow x^8=729.9\)

\(\Leftrightarrow x^8=6561\)

\(\Leftrightarrow x^8=3^8\)

\(\Leftrightarrow x=3\)

Vậy : Ở cả 2 câu x đều bằng 3 nhé.

~ Rất vui vì giúp đc bn ~

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

\(\frac{x^7}{81}\) = 27

x⁷ : 81 = 27

x⁷        = 27 . 81 = 3³ . 3⁴

x⁷         = 3⁷

Suy ra : x = 3

Bài 2 :

   \(\frac{x^8}{9}\) = 729

x⁸ : 9     = 729

x⁸          =  729 . 9

x⁸          =     3⁶ . 3²

x⁸              =              3⁸

Suy ra : x = 3

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)