a, Nghiệm = -2

b,Ngiệm = -5 và 3

c,Nghiện = -1

có cách giả không bạn

cách giải

Ta có : \(x^3-x^2-21x+45=0\)

=> \(x^3-3x^2+2x^2-6x-15x+45=0\)

=> \(x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

=> \(\left(x^2+2x-15\right)\left(x-3\right)=0\)

=> \(\left(x^2+3x-5x-15\right)\left(x-3\right)=0\)

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

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

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

Vậy phương trình có nghiệm là x = 5, x = -3, x = 3 .

a. \(x^3-x^2-21x+45=0\Rightarrow\left(x^3+5x^2\right)-\left(6x^2+30x\right)+\left(9x+45\right)=0\)

\(\Rightarrow x^2\left(x+5\right)-6x\left(x+5\right)+9\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(x-3\right)^2=0\Rightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

Vậy x=-5 hoặc x=3

b. \(2x^3-5x^2+8x-3=0\Rightarrow\left(2x^3-x^2\right)-\left(4x^2-2x\right)+\left(6x-3\right)=0\)

\(\Rightarrow x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right)\left(x^2-2x+3\right)=0\Rightarrow2x-1=0\)do \(x^2-2x+3\ne0\forall x\)

\(\Rightarrow x=\frac{1}{2}\) 

a) x³- 6x²+11x - 66 = 0

⇔x²( x - 6) + 11( x - 6) = 0

⇔( x - 6)( x² + 11 ) = 0

Do : x² + 11 > 0 ∀x

⇒ x - 6 = 0

⇒ x = 6

Vậy,...

b) x³- x²- 21x + 45=0

⇔ x³ - 3x² + 2x² - 6x - 15x + 45 = 0

⇔ x²( x - 3) + 2x( x - 3) - 15( x - 3) = 0

⇔ ( x - 3)( x² + 2x - 15 ) = 0

⇔ ( x - 3)( x² - 3x + 5x - 15 ) = 0

⇔ ( x - 3)[ x( x - 3) + 5( x - 3) ] = 0

⇔ ( x - 3)²( x + 5) = 0

⇔ x = 3 hoặc x = -5

Vậy,...

2)  2x⁴-21x³+74x²-105x+50=0

<=>(2x⁴-2x³)+(-19x³+19x²)+(55x²-55x)+(-50x+50)=0

<=>2x³.(x-1)-19x².(x-1)+55x.(x-1)-50.(x-1)=0

<=>(x-1)(2x³-19x²+55x-50)=0

<=>(x-1)[(2x³-20x²+50x)+(x²+5x-50)]=0

<=>(x-1)[2x.(x-5)²+(x²-5x+10x-50)]=0

<=>(x-1){2x.(x-5)²+[x.(x-5)+10.(x-5)]}=0

<=>(x-1)[2x.(x-5)²+(x-5)(x+10)]=0

<=>(x-1)(x-5)(2x²-10x+x+10)=0

<=>(x-1)(x-5)(2x²-5x-4x+10)=0

<=>(x-1)(x-5)[x.(2x-5)-2.(2x-5)]=0

<=>(x-1)(x-5)(x-2)(2x-5)=0

<=>x=1 hoặc x=5 hoặc x=2 hoặc x=5/2