\(\left(5-x\right)\left(9x^2-4\right)=0\)

=>\(\left(x-5\right)\left(3x-2\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\3x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\left(5-x\right)\left(9x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

a)\(4x^3-9x=0\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

Vậy x = 0 hoặc \(x=\frac{3}{2}\)

b) \(x^3+8x=0\Leftrightarrow x\left(x^2+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-8\left(L\right)\end{cases}}\)

Vậy x = 0

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

\(\Leftrightarrow\orbr{\begin{cases}-x^2+9=0\\x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=9\\x=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}\)

Vậy ...

mấy cái kia cũng làm giống vậy

1)\(x^2-x=x\left(x-1\right)=0\)

\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vì có mũ 2 nên được bỏ căn. Cái này lớp 9 mới dạy :((

(=) 2 - 9x + 2 ( 1 + x ) = 0

 (=) 2 - 9x + 2 + 2x      = 0

(=) - 9x + 2x                 = 0 - 2 - 2

(=)   - 7x                       = -4

(=)        x                       = -4 : ( -7 )

 (=)      x                        = \(\frac{4}{7}\)

Vì  \(\left(9x^2-1\right)^2\ge0;\left|x-\frac{1}{3}\right|\ge0\Rightarrow\left(9x^2-1\right)^2+\left|x-\frac{1}{3}\right|\ge0\)

Để \(\left(9x^2-1\right)^2+\left|x-\frac{1}{3}\right|=0\Leftrightarrow\hept{\begin{cases}9x^2-1=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow x=\frac{1}{3}}\)

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

Mà \(\hept{\begin{cases}x^3+x\ge0\\9x^2+9\ge0\end{cases}}\) và \(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(\Rightarrow\hept{\begin{cases}x^3+x=0\\9x^2+9=0\end{cases}}\)

Mà \(9x^2\ge0\Leftrightarrow9x^2+9>0\)

Vậy \(x\in\left\{\varnothing\right\}\)

b) \(\left(3x+2\right)-\left(x-1\right)=4\left(x+1\right)\)

\(\Leftrightarrow3x+2-x+1=4x+4\)

\(\Leftrightarrow\left(3x-x\right)+\left(2+1\right)=4x+4\)

\(\Leftrightarrow2x+3=4x+4\)

\(\Leftrightarrow2x-4x=4-3\)

\(\Leftrightarrow-2x=1\)

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

Vậy\(x=\frac{-1}{2}\)

c) \(2\left(x-1\right)-5\left(x+2\right)=-10\)

\(\Leftrightarrow2-2-5x-10=-10\)

\(\Leftrightarrow2-2-5x=0\)

\(\Leftrightarrow0-5x=0\)

\(\Leftrightarrow5x=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0

CÂU A

Để M.....>0 suy ra \(\hept{\begin{cases}x+10>0\\x-7>0\end{cases}}\)hoặc \(\hept{\begin{cases}x+10< 0\\x-7< 0\end{cases}}\) 

suy ra \(\hept{\begin{cases}x>-10\\x>7\end{cases}}\)hoặc\(\hept{\begin{cases}x< -10\\x< 7\end{cases}}\)

suy ra   x>-10  hoặc x<7                    suy ra -10<x<7    

NGOC KHONG cảm ơn bạn!! Mà bạn làm giúp mình câu b luôn đc ko vậy?? :<

(9x+ 9.9)=0
(9x+81)=0
=>>9x=0-81
9x =-81
x= -81 :9
x= -9
Vậy x=-9

cảm ơn bạn

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

\(\Leftrightarrow\left[{}\begin{matrix}x^3+x=9x^2+9\\x^3+x=-9x^2+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x^2+1\right)\left(x-9\right)=0\\\left(x^2+1\right)\left(x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)