a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

Ta có: \(A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)\)

\(=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)\)

\(=\frac{8}{3}.\left(-\frac{3}{4}\right)\)

\(=-2\)

\(B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)

\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)

\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)\)

\(\Rightarrow B=0\)

Vì -2 < 0 nên A < B

Vậy A < B

Câu a) số lớn lắm

b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)

=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)

=> \(\frac{1}{27}\cdot3^x=3^3\)

=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)

=> x = 6

b) \(\left(7x+2\right)^{-1}=3^{-2}\)

=> \(\frac{1}{7x+2}=\frac{1}{9}\)

=> 7x + 2 = 9

=> 7x = 7

=> x = 1

Bài 2:

a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)

\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)

\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)

b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)

\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)

c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)

d) Tương tự

Trả lời rõ cho mik có đc k mn

3^ x . 3^ 2= 729

3^ x  . 9        = 729

3^ x               = 729: 9

3^x                 =81 

vậy    x            = 3^3

5^ x . 625 = 3125

5^x           = 3125:625

5^x            =   5

vậy x          = 1

( 2x+1 )^ 3 = 27

( 2x+1) ^ 3 = 3^3

vậy 2x+1 =  3

  vậy x = 0         

a)(x-2)^4=4^4

=(x-2)=4 sr x=4+2

sr x=6

xin lỗi nha mình chỉ biết làm bài a thôi mong cậu thông cảm và kb với mình nhé

Mình biết bài nào làm bài đó thôi nhé

a) (x-2)⁴ = 256

=> x-2 = 4

x = 4+2

x = 6

b) \(\frac{x^4}{256}=81\)

\(\Rightarrow\frac{x^4}{4^4}=3^4\)

Từ đây có thể làm theo 2 cách khác nhau

C1 : \(\frac{x^4}{4^4}=81\)

\(\Rightarrow\left(\frac{x}{4}\right)^4=81\)

\(\Rightarrow\frac{x}{4}=3\)

x = 3.4

x = 12

C2 : \(\frac{x^4}{4^4}=3^4\)

=> x⁴ = 3⁴.4⁴

x⁴ = (3.4)⁴

x⁴ = 12⁴

<=> x = 4

c) \(125.\left(x+\frac{4}{5}\right)^3=729\)

\(\left(x+\frac{4}{5}\right)^3=729:125\)

\(\left(x+\frac{4}{5}\right)^3=5,832\)

\(\Rightarrow x+\frac{4}{5}=1,8=\frac{9}{5}\)

\(x=\frac{9}{5}-\frac{4}{5}\)

\(x=\frac{5}{5}=1\)

\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}\left(x^2+4x+4\right)-\dfrac{5}{4}\left(x^2-1\right)=\dfrac{3}{2}x\left(x-2\right)-x-4\)

\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}x^2-\dfrac{16}{3}x-\dfrac{16}{3}-\dfrac{5}{4}x^2+\dfrac{5}{4}=\dfrac{3}{2}x^2-3x-x-4\)

\(\Leftrightarrow x^2\cdot\dfrac{-25}{12}-\dfrac{25}{3}x-\dfrac{103}{12}-\dfrac{3}{2}x^2+4x+4=0\)

\(\Leftrightarrow\dfrac{-43x^2}{12x}-\dfrac{13x}{3}-\dfrac{55}{12}=0\)

\(\Leftrightarrow43x^2+52x+55=0\)

\(\text{Δ}=52^2-4\cdot43\cdot55=-6756< 0\)

Do đó: Phương trình vô nghiệm