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a) (2x + 3)2 = \(\frac{9}{121}=\left(\frac{3}{11}\right)^2=\left(-\frac{3}{11}\right)^2\)
Trường hợp 1: \(2x+3=\frac{3}{11}\)
\(2x=\frac{3}{11}-3=-\frac{30}{11}\)
\(x=-\frac{30}{11}:2=-\frac{15}{11}\)
Trường hợp 2: \(2x+3=-\frac{3}{11}\)
\(2x=-\frac{3}{11}-3=-\frac{36}{11}\)
\(x=-\frac{36}{11}:2=-\frac{18}{11}\)
Vậy \(x=-\frac{15}{11}\)hoặc \(x=-\frac{18}{11}\)
b,(3x-1)3= -8/27= (-2/3)^3
<=> 3x-1 = =2/3
<=>x=1/9 Mjk thấy phần a có bạn lm rồi nên bổ sung phần b
Chúc các bạn học tốt nhé^^

a) \(\left(2x+3\right)^2=\frac{9}{121}\)
Ta có: \(\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)
\(\Rightarrow2x+3\in\left\{\frac{3}{11};\frac{-3}{11}\right\}\)
\(\Rightarrow x\in\left\{\frac{-15}{11};\frac{-18}{11}\right\}\)
Vậy \(x\in\left\{\frac{-15}{11};\frac{-18}{11}\right\}\)
b) \(\left(3x-1\right)^3=\frac{-8}{27}\)
Ta có: \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
a.
\(\left(2x+3\right)^2=\frac{9}{121}\)
\(\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\)
\(2x+3=\pm\frac{3}{11}\)
TH1:
\(2x+3=\frac{3}{11}\)
\(2x=\frac{3}{11}-3\)
\(2x=-\frac{30}{11}\)
\(x=-\frac{30}{11}\div2\)
\(x=-\frac{15}{11}\)
TH2:
\(2x+3=-\frac{3}{11}\)
\(2x=-\frac{3}{11}-3\)
\(2x=-\frac{36}{11}\)
\(x=-\frac{36}{11}\div2\)
\(x=-\frac{18}{11}\)
Vậy \(x=-\frac{15}{11}\) hoặc \(x=-\frac{18}{11}\)
b.
\(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(3x-1=-\frac{2}{3}\)
\(3x=-\frac{2}{3}+1\)
\(3x=\frac{1}{3}\)
\(x=\frac{1}{3}\div3\)
\(x=\frac{1}{9}\)
Chúc bạn học tốt ^^


3^ x -1 = 1/243
3^x =1/243 +1
3^x = 244 / 243
Ta thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn
81^-2x . 27^x =9^5
81^-2 . 81^x . 27^x =9^5
1/9^4 . (81.27)^x =9 ^5
3^6x = 9^5 : 1/9^4
3^6x = 9^9
3^6x = 3^18
=> 6x =18
x=3
2^x +2^x +3 =144
2.(2^x) =141
2^x+1 = 141
Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn

a/ Ta có: 27x : 3x = 9
=> 33x-x = 9 =\(\left(\pm3\right)^2\)
=> 3x - x = 2
=> 2x = 2
=> x = 1
Vậy x = 1
b/ Ta có:
1/2 . 2x + 4 . 2x = 9 . 25
=> 2x . ( 1/2 + 4 ) = 9 . 32
=> 2x . 9/2 = 288
=> 2x = 64
=> x = 32
Vậy x = 32
a ) 27x : 3x = 9
<=> ( 27 : 3 )x = 9
<=> 9x = 9
=> x = 1
b )\(\frac{1}{2}.2x+4.2x=9.2^5\)
<=> x + 8x = 9.32
<=> 9x = 288
=> x = 288 : 9 = 32

\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) 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)\(\left(2x+3\right)^2=\frac{9}{121}\\ \Leftrightarrow\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{matrix}\right.\)
Vậy...
b)\(\left(3x-1\right)^3=\frac{-8}{27}\\ \Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\\ 3x-1=\frac{-2}{3}\\ \Rightarrow x=\frac{1}{9}\)
Vậy...
a) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Rightarrow2x+3=\pm\frac{3}{11}\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{3}{11}-3=-\frac{30}{11}\\2x=\left(-\frac{3}{11}\right)-3=-\frac{36}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-\frac{30}{11}\right):2\\x=\left(-\frac{36}{11}\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}.\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(\Rightarrow3x=\left(-\frac{2}{3}\right)+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}.\)
Chúc bạn học tốt!
\(9^{2x+1}=27^3\\ =>\left(3^2\right)^{2x+1}=\left(3^3\right)^3\\ =>3^{2\left(2x+1\right)}=3^{3\cdot3}\\ =>3^{4x+2}=3^9\\ =>4x+2=9\\ =>4x=9-2\\ =>4x=7\\ =>x=\dfrac{7}{4}\)
Vậy: ...
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