Question

tìm x bt:a)2x(x-3)+(3-x)=0                     c)x^4-x^2=0                e)10x-x^2-25=0

             b)3x(x+5)-6(x+5)=0                   d)4x-8x+1b=0

Answer 1

a) 2x(x-3) + (3-x) = 0

2x(x - 3) - (x - 3) = 0

(2x - 1)(x - 3) = 0

=> 2x - 1 = 0 hoặc x - 3 = 0

\(x=\dfrac{1}{2}\) hoặc x = 3

b) 3x(x+5)-6(x+5)=0

(3x - 6)(x + 5) = 0

=> 3x - 6 = 0 hoặc x + 5 = 0

x = 2 hoặc x = -5

Answer 2

`a)`

`2x(x-3)+(3-x)=0`

`<=> 2x(x-3)-(x-3)=0`

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

`b)`

`3x(x+5)-6(x+5)=0`

`<=>(x+5)(3x-6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-6=0\end{matrix}\right.\)

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

`c)`

`x^4-x^2=0`

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

 

 

Answer 3

e) 10x - x² - 25 = 0

x² - 10x + 25 = 0 (chuyển vế đổi dấu)

(x - 5)² = 0

=> x - 5 = 0

x = 5