a. x=5/6

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

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

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

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

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

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

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

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

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

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

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

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

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

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

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

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

a/\(x:27=3,6\)
\(\Rightarrow x=97,2\)
b/\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Rightarrow\left(2x+1\right)^2=81\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)

a) x=3,6\(\times\) 27=97,2

1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)

\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)

\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)

\(\Rightarrow3^x=3\)

\(\Rightarrow x=1\)

2) \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)

\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=-46\)

3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)

\(\Rightarrow2^{2x-3}=2^{49}\)

\(\Rightarrow2x-3=49\)

\(\Rightarrow2x=52\)

\(\Rightarrow x=26\)

mik cảm ơn .

b: =>(3x-1)(3x+1)(2x+3)=0

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)

c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x-1/3=19/12 hoặc 2x-1/3=-19/12

=>2x=23/12 hoặc 2x=-15/12=-5/4

=>x=23/24 hoặc x=-5/8

d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)

=>-5/6x=-3/2

=>x=3/2:5/6=3/2*6/5=18/10=9/5

e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4

=>2/5x=5/4 hoặc 2/5x=-1/4

=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8

f: =>14x-21=9x+6

=>5x=27

=>x=27/5

h: =>(2/3)^2x+1=(2/3)^27

=>2x+1=27

=>x=13

i: =>5^3x*(2+5^2)=3375

=>5^3x=125

=>3x=3

=>x=1

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 =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

1) Xét rằng x > 7 <=> A < 0

Lại xét x < 7 thì mẫu là một số nguyên dương. P/s A có tử và mẫu đều là số dương, mà tử lại bất biến

A(max) <=> mẫu 7 - x nhỏ nhất <=> 7 - x = 1 => x = 7 - 1 = 6 <=> A = 1

Từ những điều trên thì A sẽ có GTLN khi và chỉ khi x = 6

a. \(27\left(3x-1\right)^3=-\left|-8\right|\)

\(\Rightarrow27\left(3x-1\right)^3=-8\)

\(\Rightarrow\left(3x-1\right)^3=-8:27\)

\(\Rightarrow\left(3x-1\right)^3=\dfrac{-8}{27}\)

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

\(\Rightarrow3x-1=\dfrac{-2}{3}\)

\(\Rightarrow3x=\dfrac{-2}{3}+1=\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3}:3=\dfrac{1}{9}\)

Vậy...

b. \(\dfrac{2x}{-15}=\dfrac{-120}{x}\)

\(\Leftrightarrow2x.x=\left(-15\right).\left(-120\right)\)

\(\Rightarrow2.x^2=1800\)

\(\Rightarrow x^2=1800:2=900\)

\(\Rightarrow x^2=\left(\pm30\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)

Vậy...

Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

ĐKXĐ: x<>1/2

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

=>\(\left(2x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)