a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

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

a, \(\frac{1}{9}.27^n=3^n\Leftrightarrow\frac{1}{9}.3^{3.n}=3^n\Leftrightarrow\frac{1}{3^2}=3^n:3^{3n}\Leftrightarrow\frac{1}{3^2}=3^{n-3n}=3^{2n}\)

=> 3^2n . 3^2 = 1 => 3^( 2n + 2) = 3^0 => 2n + 2 = 0 => 2n = - 2 => n = - 1 

b, 3^-2.3^4 .3^n = 3^ 7 => 3^ ( -2 + 4 + n) = 3^7 => 3^ (n+ 2) = 3^7 => n + 2 = 7 => n = 5

b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)

c) \(\left(3x+2\right)^3=-27\)

\(\Rightarrow\left(3x+2\right)^3=\left(-3\right)^3\)

\(\Rightarrow3x+2=-3\)

\(\Rightarrow3x=\left(-3\right)-2\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=\left(-5\right):3\)

\(\Rightarrow x=-\frac{5}{3}\)

Vậy \(x=-\frac{5}{3}.\)

Chúc bạn học tốt!

Bạn ơi, gõ Công thức trực quan cho dễ nhìn đi bạn! :)

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

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

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

\(\Rightarrow x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}.\)

d) \(27^x:3^x=9\)

\(\Rightarrow\left(27:3\right)^x=9\)

\(\Rightarrow9^x=9\)

\(\Rightarrow9^x=9^1\)

\(\Rightarrow x=1\)

Vậy \(x=1.\)

e) \(\frac{16}{2^x}=2\)

\(\Rightarrow2^x=16:2\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

\(A=1+3+3^2+3^3+...+3^{101}\)

\(3A=3+3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\left(3^{101}-1\right):2\)

Thu gọn tổng sau:

A=1+3+3²+3³+...+3¹⁰⁰ 

B= 2¹⁰⁰-2⁹⁹-2⁹⁸-2⁹⁷-...-2²-2

C= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷-...+3²-3+1 

_​a) 3^x+2 + 5.3^x+1 = 648

=> 3^x+1.3+ 5.3^x+1= 648

=> 3^x+1(3+ 5) = 648

=> 3^x+1.8 = 648

=> 3^x+1 = 81

=>3^x+1= 3⁴

=> x+1 = 4

=> x = 3

vậy x= 3

a)1/9.27^n=3^n

             3^n=3^n

         =>n={0;1;2;3;...}     

a) n= 2;3;5;7;...(n là số nguyên)

Mk làm lun, ko viết lại đề bài nữa nhé =))

a) \(\Leftrightarrow\)\(3^2.3^{n+1}=9^4\)

\(\Leftrightarrow3^{n+1}=9^4:3^2\)

\(\Leftrightarrow3^{n+1}=3^6\)

\(\Rightarrow n+1=6\)

\(\Leftrightarrow n=6-1\)

\(\Rightarrow n=5\)

b)\(\Leftrightarrow2^n.\left(\frac{1}{2}+4\right)=9.2^5\)

\(\Leftrightarrow2^n.\frac{9}{2}=9.2^5\)

\(\Rightarrow2^n=\left(9.2^5\right):\frac{9}{2}\)

\(\Rightarrow2^n=468:\frac{9}{2}\)

Tự tính nốt KQ giúp mk nha ♥

a)1/9*3⁴*3ⁿ⁺¹=9⁴

3⁴*3ⁿ⁺¹ =9⁴/(1/9)=9⁴*9=9⁵

3⁴*3ⁿ⁺¹=3⁴⁺ⁿ⁺¹=9⁵=(3²)⁵=3¹⁰

=>4+n+1=10

n=10-4-1

Vậy n=5

b)1/2*2ⁿ+4*2ⁿ=9*2⁵

2ⁿ*(1/2+4)=9*2⁵

2ⁿ*4.5=9*2⁵

2ⁿ=9*2⁵/4.5=2⁵*2=2⁶

=> Vậy n=6

\(a)\frac{1}{9}.3^4.3^{n+1}=9^4\)

\(\Rightarrow3^{-2}.3^4.3^{n+1}=3^8\)

\(\Rightarrow3^{-2+4+n+1}=3^{n+3}=3^8\)

\(\Rightarrow n+3=8\Rightarrow n=5\)

\(b)\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n\left(\frac{1}{2}+4\right)=9.2^5\)

\(\Rightarrow2^n.\frac{9}{2}=9.2^5\Rightarrow2^n=9.2^5.\frac{2}{9}=\frac{9.2^5.2}{9}=2^6\)

\(\Rightarrow n=6\)