a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)

\(\Leftrightarrow x^3-27-x^3+3x=0\)

\(\Leftrightarrow x=9\)

b: Ta có: \(8x^4+x=0\)

\(\Leftrightarrow x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

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: \(\left(2x-1\right)^2-25=0\)

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

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

1) ĐKXĐ: \(x\notin\left\{-2;2\right\}\)

Ta có: \(\dfrac{x-1}{x+2}-\dfrac{9}{x^2-4}=\dfrac{-3}{x-2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2-3x+2-9=-3x-6\)

\(\Leftrightarrow x^2-3x-7+3x+6=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={1;-1}

2)

Sửa đề: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

\(\Leftrightarrow\dfrac{3x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

Suy ra: \(3x-3-x-3=x^2-3x+x-3\)

\(\Leftrightarrow x^2-2x-3=2x-6\)

\(\Leftrightarrow x^2-2x-3-2x+6=0\)

\(\Leftrightarrow x^2-4x+3=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={1}

`1)(x-1)/(x+2)-9/(x^2-4)=-3/(x-2)(x ne 2)`

`<=>x^2-3x+2-9=-3x-6`

`<=>x^2-1=0`

`<=>x=+-1`

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

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

a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)

b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)

c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)