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

từ đó suy ra 

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

2x=3/11-3

2x=-2/8/11

x=-2/8/11:2

x=-1/4/11

b,

(3x-1)^3=-8/27

(3x-1)^3=(-2/3)^3

Vậy suy ra 

3x-1=-2/3

3x=-2/3+1

3x=1/3

x=1/3:3

x=1/9

Ko thấy x             

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!

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

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

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

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

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

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) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

câu b thiếu bn ơi

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

a: Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

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....