Tìn x ạ?

-1 5/27-(3x-7/9)3=-24/27

 (3x-7/9)3 =-32/27-(-24/27)

 (3x-7/9)3=-8/27 

(3x-7/9)3=(-2/3)3

⇒3x-7/9=-2/3

  3x =-2/3+7/9

 3x =1/9

x =1/9:3

x =1/27

Chúc bạn học tốt!

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báo cáo

báo cáo nha NGÔ QUANG LÍ

-1 5/27-(3x-7/9)³=-24/27

           (3x-7/9)³ =-32/27-(-24/27)

           (3x-7/9)³=-8/27

           (3x-7/9)³=(-2/3)³

⇒3x-7/9=-2/3

   3x      =-2/3+7/9

   3x      =1/9

     x      =1/9:3

     x      =1/27

Chúc bạn học tốt!

a) xem lại đề

b) 3^x-1=27

=>3^x-1=3³

=>x-1=3

=>x=3+1

=>x=4

c)3^x+1=9

=>3^x+1=3²

=>x+1=2

=>x=2-1

=>x=1

a) x²=x³

⇒ x²-x³=0

⇒ x²-x².x=0

⇒ x².(x+1)=0

⇒\(\left\{{}\begin{matrix}\text{x}^2=0\\\text{x}+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\text{x}=0\\\text{x}=0+1=1\end{matrix}\right.\)

⇒ \(\text{x}\in\left\{0;1\right\}\)

b) 3^x-1=27

⇒ 3^x-1=3³

⇒ x-1=3

⇒ x= 3+1=4

c) 3^x+1=9

⇒ 3^x+1= 3³

⇒ x+1=3

⇒ x=3-1=2

-1 . \(\dfrac{5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = -\(\dfrac{24}{27}\)

\(\Rightarrow\) \(\dfrac{-5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-24}{27}\)

\(\Rightarrow\) \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-5}{27}\) - \(\dfrac{-24}{27}\) = \(\dfrac{19}{27}\)

\(\Rightarrow\) \(3^3\) . \(\dfrac{7^3}{3^6}\) = \(\dfrac{19}{3^3}\)

\(\Rightarrow\) \(\dfrac{x^3.7^3}{3^3}\) = \(\dfrac{19}{3^3}\)

\(\Rightarrow\) \(x^3\) . \(7^3\) = 19

3x+27=9

=>3x= 9-27

=> 3x= -18

=> x= -6

3x+27=9

x        =(9-27):3

x       =-6

3x + 27 = 9

3x         = 9 - 27

3x         = 9 + (-27)

3x         =    -18

 x          = -18 : 3

x           =   -6 \(\in\)Z

Vậy x  = -6

k mk nha bn

=> [3x-7/9]^3= -8/27

=> 3x -7/9 = -2/3 => 3x = 1/9=> x= 1/27

đúng cái nhé

1/

\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)

\(\Leftrightarrow43x=0\Rightarrow x=0\)

2/

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)

\(\Leftrightarrow x^3+27-x^3+x=27\)

\(\Leftrightarrow x=0\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.9=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+9x=27\\ 9x+27=27\\ 9x=0\\ x=0\)