Question

giải các pt sau 

a\   \(2x^2-x=3-6x\)

b\   \(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right)x^2\)

Answer 1

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

\(\Leftrightarrow2x^2-x-3x+6x=0\)

\(\Leftrightarrow2x^2+2x=0\)

\(\Leftrightarrow2x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

b) \(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right)x^2\)

\(\Leftrightarrow x^2-3x+5=x^2\)

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

<=> -3x+5=0

<=> \(x=\frac{5}{3}\)

Answer 2

a) 2x^2 - x = 3 - 6x

<=> 2x^2 - x - 3 + 6x = 0

<=> 2x^2 + 5x - 3 = 0

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

<=> 2x - 1 = 0 hoặc x + 3 = 0

<=> 2x = 0 + 1 hoặc x = 0 - 3

<=> 2x = 1 hoặc x = -3

<=> x = 1/2 hoặc x = -3

b) (x + 2)(x^2 - 3x + 5) = (x + 2)x^2

<=> x^3 - 3x^2 + 5x + 2x^2 - 6x + 10 = x^3 + 2x^2

<=> x^3 - 3x^2 + 5x + 2x^2 - 6x + 10 - x^3 - 2x^2 = 0

<=> 3x^2 + x - 10 = 0 (đổi dấu)

<=> 3x^2 + 6x - 5x - 10 = 0

<=> 3x(x + 2) - 5(x + 2) = 0

<=> (3x - 5)(x + 2) = 0

<=> 3x - 5 = 0 hoặc x + 2 = 0

<=> 3x = 0 + 5 hoặc x = 0 - 2

<=> 3x = 5 hoặc x = -2

<=> x = 5/3 hoặc x = -2