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 ^^

a) x = 4                

b) x = 7     

c) x = 2                

d) x = 5

e) x = 2                

f) x= 1. 

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a) x = 4

b) x = 7

c) x = 2

d) x = 5

e) x = 2

 f) x= 1

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) Vì \(\left(2.x+3\right)^2=\dfrac{9}{121}\Rightarrow\left\{{}\begin{matrix}2.x+3=\dfrac{3}{11}\\2.x+3=-\dfrac{3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{11}\\x=-\dfrac{18}{11}\end{matrix}\right.\)

b) Vì \(\left(3.x-1\right)^3=-\dfrac{8}{27}\Rightarrow3.x-1=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{9}\)

a)  \(\left(2x+3\right)^2=\frac{9}{21}\)

<=>  \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-1\frac{4}{11}\\x=-1\frac{7}{11}\end{cases}}\)

Vậy...

b)  \(\left(3x-1\right)^3=\frac{-8}{27}\)

<=>   \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)

<=>   \(3x-1=\frac{-2}{3}\)

<=>  \(3x=\frac{1}{3}\)

<=>  \(x=\frac{1}{9}\)

Vậy....

a) (2x + 3)² = 9/121 

Ta có: 9/121 = (3/11)² = (-3/11)²

=> 2x + 3 thuộc {3/11; -3/11}

=> x thuộc {-15/11; -18/11}

b) (3x - 1)³ = -8/27 = (-2/3)³

=> 3x - 1 = -2/3 

=> x = 1/9 

\(\left(2x+3\right)^2=\frac{9}{121}\)

\(\Rightarrow\left(2x+3\right)^2=\hept{\begin{cases}\left(\frac{3}{11}\right)^2\\\left(\frac{-3}{-11}\right)^2\end{cases}}\)

\(\Rightarrow2x+3=\hept{\begin{cases}\frac{3}{11}\\\frac{-3}{-11}\end{cases}}\)

a)\(\left(2x+3\right)^2=\frac{9}{121}\)

\(\Leftrightarrow\left(2x+3\right)^2=\frac{3^3}{11^2}\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)

\(\Leftrightarrow\hept{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-\frac{30}{11}\Leftrightarrow x=-\frac{15}{11}\\2x=-\frac{36}{11}\Leftrightarrow x=-\frac{18}{11}\end{cases}}}\)

b)\(\left(3x-1\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-1\right)^3=\sqrt[3]{\left(-\frac{8}{27}\right)}\)

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

\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)