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 \(-\frac{18}{11}\)

Tìm a: 1/2a=5/15+3/4=1/3+3/4=13/12

=> a=13/12:1/2=13/12.2=13/6

Tìm x: -2/3x=18/15-0,2=6/5-0,2=1

=>x=1:-2/3=-3/2

cho mk nha

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

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

\(\Rightarrow3x-1=\frac{2}{3}\)

\(\Rightarrow3x=\frac{5}{3}\)

\(\Rightarrow x=\frac{5}{3}:3\)

\(\Rightarrow x=\frac{5}{9}\)

tíc mình nha

3x=8/27+1

3x=35/27

x=35/27:3

x=35/81

x = 5 / 9

1.

| x + 2 | = | 2 - 3x |

xét 2 trường hợp :

+) TH1 :

2 - 3x = x + 2

-3x + x = 2 + 2

2x = 4

x = 4 : 2 = 2

+) TH2 : 

2 - 3x = - ( x + 2 )

2 - 3x = -x - 2

-3x - x = 2 - 2

-4x = 0

x = 0 : ( -4 )

x = 0

bài còn lại tương tự

tìm x,biết:(x^2-3)^2=16

Bài này không phù hợp với lớp 6 bạn à.

(3.x)^2 : 3^3=243

9.x^2 : 27=243

9.x^2=243.27

9.x^2=6561

x^2=6561:9

x^2=729

X=27

\(\left(x-1\right).3+3x\left(x-4\right)+1=0\)

\(\Rightarrow3x-3+3x^2-12x+1=0\)

\(\Rightarrow3x^2-9x-2=0\)

\(\Rightarrow3\left(x^2-\frac{2.3}{2}.x+\frac{9}{4}\right)-\frac{35}{4}=0\)

\(\Rightarrow3\left(x-\frac{3}{2}\right)^2=\frac{35}{4}\Rightarrow\left(x-\frac{3}{2}\right)^2=\frac{35}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{2}=\sqrt{\frac{35}{12}}\\x-\frac{3}{2}=-\sqrt{\frac{35}{12}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{35}{12}}+\frac{3}{2}\\x=\frac{3}{2}-\sqrt{\frac{35}{12}}\end{cases}}\)

Vậy.....................

2(x - 3) - (3x - 5) = x + 20 - (x - 1)

=> 2x - 6 - 3x + 5 = x + 20 - x + 1

=> -x - 1 = 21

=> -x = 21 + 1

=> -x = 22

=> x= 22

\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow9x^3-18x^2+9x-29=0\)

\(\Leftrightarrow x=2,208024627\)