Question

bài 1 ; tìm x

a) x (x - 3 ) - 2x + 6 = 0

b) (3x - 5 ) ( 5x-7 ) + (5x +1) (2-3x) =4

Answer 1

a: x(x-3)-2x+6=0

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

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

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

b: \(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)

=>\(15x^2-21x-25x+35+10x-15x^2+2-3x=4\)

=>-39x=4-35=-31

=>\(x=\dfrac{31}{39}\)

Answer 2

`a)x(x-3)-2x+6=0`

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

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

`=>(x-3)(x-2)=0=>[(x=3),(x=2):}`

`b)(3x-5)(5x-7)+(5x+1)(2-3x)=4`

`=>15x^2-21x-25x+35+10x-15x^2+2-3x=4`

`=>39x=33`

`=>x=11/13`

Answer 3

Ta có : x(x - 3) - 2x + 6 = 0

<=> x(x - 3) - (2x - 6) = 0

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

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

⇔\orbr{𝑥−2=0𝑥−3=0⇔\orbr{x−2=0x−3=0​

⇔\orbr{𝑥=2𝑥=3⇔\orbr{x=2x=3​

 

Answer 4