a, \(\left(2x^3+3\right)^2=\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)

Nếu \(2x+3=\frac{3}{11}\Rightarrow x=-\frac{15}{11}\)

Nếu \(2x+3=-\frac{3}{11}\Rightarrow x=-\frac{18}{11}\)

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

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

a, (2x+3)^2 = 9/121

=> 2x+3 = \(\sqrt{\frac{9}{121}}\)= \(\frac{3}{11}\)

=>x= \(\frac{\frac{3}{11}-3}{2}\) = \(-\frac{15}{11}\)

b,(3x-1)\(^3\)= \(-\frac{8}{27}\)

=> \(3x-1=\sqrt[3]{-\frac{8}{27}}=-\frac{2}{3}\)

=>\(x=\frac{-\frac{2}{3}+1}{3}=\frac{1}{9}\)

 t còn thiếu 1 nghiệm nữa là \(-\frac{18}{11}\)

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

=> 3x -1 = 2/3

3x = 5/3

x = 5/9

học tốt ^^

\(\left(x^4\right)^2=x^{12-5}\)

\(x^8-x^7=0\)

\(x^7\cdot x-x^7=0\)

\(x^7\cdot\left(x-1\right)=0\)

+) x^7 = 0 => x = 0

+) x -1 = 0 => x = 1

Vậy,...........

học tốt ^^

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\Rightarrow\left(x^4\right)^2x^5=\frac{x^{12}}{x^5}x^5\Rightarrow x^{13}=x^{12}\)

\(\Rightarrow x=1\)

a) \(\left(2x+3\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+3\right)^2=\left(\frac{3}{5}\right)^2\) hoặc \(\left(2x+3\right)^2=\left(-\frac{3}{5}\right)^2\)

\(\Rightarrow2x+3=\frac{3}{5}\) hoặc \(2x+3=-\frac{3}{5}\)

\(\Rightarrow2x=-\frac{12}{5}\) hoặc \(2x=-\frac{18}{5}\)

\(\Rightarrow x=-\frac{6}{5}\) hoặc \(x=-\frac{9}{5}\)

Vậy.......

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

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

\(\Rightarrow3x-1=\frac{1}{3}\)

\(\Rightarrow3x=\frac{4}{3}\)

\(\Rightarrow x=\frac{4}{9}\)

Vậy.......

a) (2x+3)²=\(\frac{9}{25}\)

2x+3=\(\sqrt{\frac{9}{25}}\)=\(\frac{3}{5}\)

2x=\(\frac{-12}{5}\)

x=\(\frac{-6}{5}\)

b) (3x-1)³=\(\frac{-1}{27}\)

3x-1=\(\frac{-1}{3}\)

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

x=\(\frac{2}{9}\)

a) ( 2x + 3 )² = \(\frac{9}{25}\)

\(\Rightarrow2x+3=\frac{3}{5}\)hoặc \(-\frac{3}{5}\)

\(\Rightarrow2x=-\frac{12}{5}\)hoặc \(-\frac{18}{5}\)

\(\Rightarrow x=-\frac{6}{5}\)hoặc\(-\frac{9}{5}\)

b) ( 3x - 1)³ = \(-\frac{1}{27}\)

\(\Rightarrow3x-1=-\frac{1}{3}\)

\(\Rightarrow3x=\frac{2}{3}\)

\(\Rightarrow x=\frac{2}{9}\)

1) ta có: \(x:3=y.15\Rightarrow x\cdot\frac{1}{3}=y.15\Rightarrow\frac{x}{15}=\frac{y}{\frac{1}{3}}\)

ADTCDTSBN

...

2) bn ghi thiếu đề r

3) ta có: \(3x=7y\Rightarrow\frac{x}{7}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=7k\\y=3k\end{cases}}\)

mà xy = 189 => 7k.3k = 189

                          21 k² = 189

                                 k² = 9 = 3² = (-3)² => k = 3 hoặc k  = - 3

TH1: k = 3

x = 7.3 => x  = 21

y = 3.3 => y = 9

...

4) ta có: \(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\)

ADTCDTSBN

...

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

\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)

a) Ta có: \(3x\left(x+1\right)-2x\left(x+20\right)=-1-x\)

\(\Leftrightarrow3x^2+3x-2x^2-40x+1+x=0\)

\(\Leftrightarrow x^2-36x+1=0\)

\(\Leftrightarrow x^2-36x+324-323=0\)

\(\Leftrightarrow\left(x-18\right)^2=323\)

\(\Leftrightarrow\left[{}\begin{matrix}x-18=\sqrt{323}\\x-18=-\sqrt{323}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18+\sqrt{323}\\x=18-\sqrt{323}\end{matrix}\right.\)

Vậy: \(x\in\left\{18+\sqrt{323};18-\sqrt{323}\right\}\)

b) Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)-16=0\)

\(\Leftrightarrow6x^2+19x-7-6x^2-x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

hay x=1

Vậy: x=1

c) Ta có: \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)-8=0\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3-8=0\)

\(\Leftrightarrow-4x-5=0\)

\(\Leftrightarrow-4x=5\)

hay \(x=\frac{-5}{4}\)

Vậy: \(x=\frac{-5}{4}\)