x³ - 9x - 5x² + 45 = 0

⇔ ( x³ - 5x² ) - ( 9x - 45 ) = 0

⇔ x²( x - 5 ) - 9( x - 5 ) = 0

⇔ ( x - 5 )( x² - 9 ) = 0

⇔ ( x - 5 )( x - 3 )( x + 3 ) = 0

⇔ x - 5 = 0 hoặc x - 3 = 0 hoặc x + 3 = 0

⇔ x = 5 hoặc x = ±3

\(x^3-9x-5x^2+45=0\)   

\(x^3-5x^2-9x+45=0\)   

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

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

\(\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)   

\(\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

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

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

\(x^3+6x^2+9x=0\)

\(x\left(x^2+6x+9\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}x=0\\x=-3\end{cases}}\)

x³-5x²-9x+45=0

=>(x³-5x²)-(9x-45)=0

=>x²(x-5)-9(x-5)=0

=>(x²-9)(x-5)=0

=>x²-9=0  =>x²=9  => x=3;-3

     x-5=0  =>x=5

help me

a)
(3x+1)^2 -9x (x-1) =46
9x²+6x+1-9x²+9x = 46
15x+1=46
15x=45
x=3
b)
5x (x-3) = 7 (3-x)
5x ( x - 3 ) - 7 ( 3 - x ) = 0
5x ( x - 3 ) + 7 ( x - 3 ) = 0
(5x + 7 ) ( x - 3 ) = 0
5x+7=0 hoặc x-3 = 0 
5x=-7    hoặc x=3
  x= -7 / 5 hoặc x=3

\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

a)\(7x\left(x-2\right)=\left(x-2\right)\)

\(\Leftrightarrow7x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(7x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7x-1=0\\x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=2\end{matrix}\right.\)

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

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

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

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

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

c)\(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+9x+5x^2+45=0\)

\(\Leftrightarrow x\left(x^2+9\right)+5\left(x^2+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2+9\right)=0\)

Dễ thấy: \(x^2+9\ge 9 >0\forall x\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

d,e tương tự

giúp em thé các anh chi em trên thế giới

a. x³ + 5x² +9x + 45 = 0

<=> x²(x + 5) + 9(x + 5) = 0

<=> (x + 5)(x² +9)=0

(x+5)= 0 hoặc (x² + 9)=0 (vô lý)

<=> x = -5