Question

tìm x, biết:

a) (3x-1)²+(x+3)³-5(2x-3)(x+3)=0

b) x²-4x+4-(x+3)(x-3)=0

c) x²-3x+2=0

Answer 1

c) x² - 3x + 2 = 0

x² - x - 2x + 2 = 0

(x² - x) + ( -2x+ 2) =0

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

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

\(\Rightarrow\) x-1 = 0 hoặc x-2 = 0

+) x - 1= 0\(\Rightarrow\) x= 1

+) x - 2= 0 \(\Rightarrow\) x = 2

Vậy: x = 1 hoặc x = 2

Answer 2

a: \(\Leftrightarrow9x^2-6x+1+x^2+6x+9-5\left(2x^2+6x-3x-9\right)=0\)

=>\(10x^2+10-10x^2-15x+45=0\)

=>-15x+55=0

=>x=11/3

b: \(\Leftrightarrow x^2-4x+4-x^2+9=0\)

=>13-4x=0

=>x=13/4