Question

Tìm x : 

a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49

b) 16x^2 - ( 4x - 5 )^2 = 15

c) ( 2x + 1 )^2 - ( x - 1)^2 = 0

Answer 1

a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49

=>4x²+12x+9-4x²+4=49

 =>12x+13=49

=>12x=36

=>x=3

b) 16x^2 - ( 4x - 5 )^2 = 15

=>16x²-16x²+40x-25=15

=>40x-25=15

=>40x=40

=>x=1

c) ( 2x + 1 )^2 - ( x - 1)^2 = 0

=>4x²+4x+1-x²+2x-1=0

=>3x²+6x=0

=>3x(x+2)=0

=>3x=0 hoặc x+2=0

=>x=0 hoặc x=-2

 

 

Answer 2

a) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ =>4x^2+12x+9-4x^2+4=49\\=>12x+13=49\\ =>12x=36\\ =>x=3\)

b) \(16x^2-\left(4x-5\right)^2=15\\ =>16x^2-16x^2+40x-25=15\\ =>40x-25=15\\ =>40x=40\\ =>x=1\)

c) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\\ =>4x^2+4x+1-x^2+2x-1=0\\ =>3x^2+6x=0\\ =>3x\left(x+2\right)=0\\ =>\left[\frac{3x=0}{x+2=0}\right]=>\left[\frac{x=0}{x=-2}\right]\)