Question

giúp em vs ạ

Answer 1

`a, x^2-4x-5=0`

`<=>x^2+x-5x-5=0`

`<=> (x^2+x)-(5x+5)=0`

`<=> x(x+1)-5(x+1)=0`

`<=>(x+1)(x-5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

`b, 5x^2-9x-2=0`

`<=> 5x^2+x-10x-2=0`

`<=> (5x^2+x)-(10x+2)=0`

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=2\end{matrix}\right.\)

`c, x^3+2x-12=0`

`<=> x^3+2x-8-4=0`

`<=> (x^3-8)+(2x-4)=0`

`<=> (x-2)(x^2 + 2x+4) + 2(x-2)=0`

`<=>(x-2)(x^2+2x+4+2)=0`

`<=>(x-2)(x^2+2x+6)=0`

`<=>x-2=0`

`<=>x=2`

`d, -2x^3 +x^2-3=0`

`<=> -(2x^3-x^2+3)=0`

`<=> -(2x^3 -3x^2 +3x+2x^2 -3x+3)=0`

`<=>-[(2x^3 -3x^2 +3x) +(2x^2-3x+3)]=0`

`<=> - [x(2x^3 -3x^2 +3x) +(2x^2-3x+3)]=0`

`<=> -(2x^3-3x^2+3x)(x+1)=0`

`<=>x+1=0`

`<=>x=-1`

 

 

Answer 2

\(a,x^2-4x-5=0\\ \Leftrightarrow x^2+x-5x-5=0\\ \Leftrightarrow x\left(x+1\right)-5\left(x+1\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\\ b,5x^2-9x-2=0\\ \Leftrightarrow5x^2-10x+x-2=0\\ \Leftrightarrow5x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(5x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=2\end{matrix}\right.\)

Answer 3

c) x³ + 2x - 12 = 0

x³ - 2x² + 2x² - 4x + 6x - 12 = 0

(x³ - 2x²) + (2x² - 4x) + (6x - 12) = 0

x²(x - 2) + 2x(x - 2) + 6(x - 2) = 0

(x - 2)(x² + 2x + 6) = 0

⇒ x - 2 = 0 (vì x² + 2x + 6 > 0 với mọi x ∈ R)

x = 0 + 2

x = 2

Vậy x = 2

d) -2x³ + x² - 3 = 0

-2x³ - 2x² + 3x² - 3 = 0

(-2x³ - 2x²) + (3x² - 3) = 0

-2x²(x + 1) + 3(x² - 1) = 0

-2x²(x + 1) + 3(x + 1)(x - 1) = 0

(x + 1)(-2x² + 3x - 3) = 0

⇒ x + 1 = 0 (vì -2x² + 3x - 3 < 0 với mọi x ∈ R)

x = -1

Vậy x = -1

Answer 4

g) (x² + 1)² - 5(x² + 1) + 6 = 0

Đặt m = x² + 1 (m > 0), ta có:

m² - 5m + 6 = 0

m² - 2m - 3m + 6 = 0

(m² - 2m) - (3m - 6) = 0

m(m - 2) - 3(m - 2) = 0

(m - 2)(m - 3) = 0

⇒ m - 2 = 0 hoặc m - 3 = 0

*) m - 2 = 0

m = 2 (nhận)

*) m - 3 = 0

m = 3 (nhận)

* Với m = 2

⇒ x² + 1 = 2

⇒ x² = 2 - 1

⇒ x² = 1

⇒ x = 1 hoặc x = -1

* Với m = 3

⇒ x² + 1 = 3

⇒ x² = 3 - 1

⇒ x² = 2

⇒ x = -√2 hoặc x = √2

Vậy x = -√2; x = -1; x = 1; x = √2

Answer 5

h) (x² + 6x)² - 2(x + 3)² - 17 = 0

x⁴ + 12x³ + 36x² - 2x² - 12x - 18 - 17 = 0

x⁴ + 12x³ + 34x² - 12x - 35 = 0

x⁴ + x³ + 11x³ + 11x² + 23x² + 23x - 35x - 35 = 0

(x⁴ + x³) + (11x³ + 11x²) + (23x² + 23x) - (35x + 35) = 0

x³(x + 1) + 11x²(x + 1) + 23x(x + 1) - 35(x + 1) = 0

(x + 1)(x³ + 11x² + 23x - 35) = 0

(x + 1)(x³ - x² + 12x² - 12x + 35x - 35) = 0

(x + 1)[(x³ - x²) + (12x² - 12x) + (35x - 35)] = 0

(x + 1)[x²(x - 1) + 12x(x - 1) + 35(x - 1)] = 0

(x + 1)(x - 1)(x² + 12x + 35) = 0

(x + 1)(x - 1)(x² + 5x + 7x + 35) = 0

(x + 1)(x - 1)[(x² + 5x) + (7x + 35)] = 0

(x + 1)(x - 1)[x(x + 5) + 7(x + 5)] = 0

(x + 1)(x - 1)(x + 5)(x + 7) = 0

x + 1 = 0 hoặc x - 1 = 0 hoặc x + 5 = 0 hoặc x + 7 = 0

*) x + 1 = 0

x = -1

*) x - 1 = 0

x = 1

*) x + 5 = 0

x = -5

*) x + 7 = 0

x = -7

Vậy x = -7; x = -5; x = -1; x = 1