Question

tìm x biết:

a) x(x+1)-x^2=3

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

Answer 1

`x(x+1)-x^2 = 3`

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

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

`=> x = 3`

Vậy ...

`x(3x - 5) = (3x - 5)^2`

`=> 3x^2 - 5x = 9x^2 - 30x + 25`

`=> 9x^2 - 3x^2 - 30x + 5x + 25 = 0`

`=> (9x^2 - 3x^2)- (30x - 5x) + 25 = 0`

`=> 6x^2 - 25x + 25 = 0`

`=> (3x - 5)(2x - 5) = 0`

`=> x = 5/3` hoặc `x = 5/2`

Vậy ....

Answer 2

a) \(x\left(x+1\right)-x^2=3\Leftrightarrow x^2+x-x^2=3\Leftrightarrow x=3\)

b) \(x\left(3x-5\right)=\left(3x-5\right)^2\Leftrightarrow\left(3x-5\right)\left(-2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\)

Answer 3

`x(x+1) - x^2 = 3`

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

`=> x = 3`

Vậy `x =3`

`x(3x - 5) = (3x - 5)^2`

`=> x(3x - 5) - (3x - 5)^2 = 0`

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

Th1 :

`3x - 5 = 0`

`=> 3x = 5`

`=> x = 5/3`

TH2:

`-2x + 5 =0`

`=> -2x = -5`

`=>   x = 5/2`

Answer 4

a: \(x\left(x+1\right)-x^2=3\)

=>\(x^2+x-x^2=3\)

=>x=3

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

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

=>(3x-5)(3x-5-x)=0

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

=>\(\left[{}\begin{matrix}3x-5=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\)