a) ...

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

=) 3x-2=(-3)

  3x=(-1)

   x=(-1/3)

DUYỆT CHO MÌH ĐI RỒI MÌH LẠI GIẢI TIẾP CHO

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

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

=> 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

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

a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)

\(\Leftrightarrow\frac{1}{6}\cdot6^x+6^x\cdot6^2=6^{10}+6^7\)

\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(6^3+1\right)\)

\(\Leftrightarrow6^{x-1}=6^7\Leftrightarrow x-1=7\)

\(\Leftrightarrow x=8\)

b, \(\left(\frac{1}{2}-\frac{1}{6}\right)\cdot3^{x+4}-4\cdot3^x=3^{16}-4\cdot3^{13}\)

\(\Leftrightarrow\frac{1}{3}\cdot3^{x+4}-4\cdot3^x=3^{13}\left(3^3-4\right)\)

\(\Leftrightarrow3^x\cdot3^3-4\cdot3^x=3^{13}\left(3^3-4\right)\)

\(\Leftrightarrow3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)

\(\Leftrightarrow3^x=3^{13}\Leftrightarrow x=13\)

a. x=8

b. x=13

còn cách tính thì mình quên rồi vì minh học cái này lâu lắm rồi ko nhớ đc.

a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)

\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)

\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)

\(6^{x-1}.217=6^{15}.217\)

\(6^{x-1}=6^{15}\)

\(x-1=15\)

\(x=16\)

b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)

\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)

\(3^x=3^{13}\)

\(x=13\)

\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)

=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)

=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)

=> \(3^x.3^4-4.3^x=-386339074,3\)

=> \(3^x.\left(3^4-4\right)=-386339074,3\)

=> \(3^x.77=-386339074,3\)

=> \(3^x=-386339074,3:77\)

=> \(3^x=-5017390,575\)

=> x = ... chắc tự ngồi tính đc

cái a tương tự thôi

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(5^{x+3}\left(5-3\right)=2.5^{11}\)

\(5^{x+3}.2=2.5^{11}\)

\(5^{x+3}=5^{11}\)

\(x+3=11\)

\(x=8\)

\(4^{x+3}-3.4^{x+1}=13.4^{11}\)

\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)

\(4^{x+1}.13=13.4^{11}\)

\(4^{x+1}=4^{11}\)

\(x+1=11\)

\(x=10\)

ĐÉOOOOOOOOOOOOOO!!!!!!!!!!

Hey bợn Ami 

Có fb ko ib lw

a, Ta có \(2.3^{x+2}+4.3^{x+1}=3^6.10\)

\(\Rightarrow2.3.3^{x+1}+4.3^{x+1}=3^6.10\)

\(\Rightarrow3^{x+1}.\left(6+4\right)=3^6.10\)

\(\Rightarrow3^{x+1}.10=3^6.10\)

\(\Rightarrow3^{x+1}=3^6\)

\(\Rightarrow x+1=6\)

\(\Rightarrow x=5\)

b,\(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{16}\)

\(\Rightarrow\frac{1}{2}.2^{x+4}-2^x=2^{13}.\left(1-2^3\right)\)

\(\Rightarrow2^{x+3}-2^x=2^{13}.\left(1-2^3\right)\)

\(\Rightarrow2^x.\left(2^3-1\right)=2^{13}.\left(1-2^3\right)\)

\(\Rightarrow2^x.\left(2^3-1\right)=-2^{13}.\left(2^3-1\right)\)

\(\Rightarrow2^x=2^{-13}\)

\(\Rightarrow x=-13\)

A ) 2 . 3^x+2 + 4 . 3³⁺¹ = 3⁶ . 10

      2 . 3^x . 9 + 4 . 3^x . 3 = 729 .10

         18 . 3^x + 12 . 3^x = 243 . 3 . 10

                        30 . 3^x = 243 . 30

                               3^x = 243

                                 x = 5 

\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)

d, Điều kiện: x > 1

\(log\left(x-1\right)< 0\\ \Leftrightarrow x-1< 1\\ \Leftrightarrow1< x< 2\)

e, Điều kiện: \(x>\dfrac{1}{2}\)

\(log_{\dfrac{1}{5}}\left(2x-1\right)\ge log_{\dfrac{1}{5}}\left(x+3\right)\\ \Leftrightarrow2x-1\ge x+3\\ \Leftrightarrow x\ge4\)

f, Điều kiện: x > 4

\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)