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) (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^^

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

đề yêu cầu tìm x hả

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

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

\(\Rightarrow2x+3=\frac{3}{11}\text{ hoặc }2x+3=\frac{-3}{11}\)

\(2x=\frac{3}{11}-3\text{ hoặc }2x=\frac{-3}{11}-3\)

\(2x=\frac{3}{11}-\frac{33}{11}\text{ hoặc }2x=\frac{-3}{11}-\frac{33}{11}\)

\(2x=\frac{-30}{11}\text{ hoặc }2x=\frac{-36}{11}\)

\(x=\frac{-30}{11}:2\text{ hoặc }x=\frac{-36}{11}:2\)

\(x=\frac{-30}{11}.\frac{1}{2}\text{ hoặc }x=\frac{-36}{11}.\frac{1}{2}\)

\(x=\frac{-15}{11}\text{ hoặc }x=\frac{-18}{11}\)

b: \(\Leftrightarrow x-15-27-x+x-13=-1\)

\(\Leftrightarrow x-55=-1\)

hay x=54

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

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!