Question

Tìm x biết:

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

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

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

Answer 1

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

=>\(9\text{x}^2+24x+16-9\text{x}^2+1\)\(=49\)

=>\(24\text{x}+17=49\)

=> 24x = 32

=> x = \(\dfrac{4}{3}\)

b) \(\left(3\text{x}-1\right)^2-\left(3\text{x}-2\right)^2=0 \)

\(=>9\text{x}^2-6\text{x}+1-9\text{x}^2+12\text{x}-4=0\)

\(=>6\text{x}-3=0\)

=> 6x = 3

=> x = \(\dfrac{1}{2}\)

c) \(\left(2\text{x}+1\right)^2-\left(x-1\right)^2=0\)

\(=>4\text{x}^2+4\text{x}+1-x^2+2\text{x}-1=0\)

=> \(3\text{x}^2+6\text{x}=0\)

=> \(3\text{x}\left(x+2\right)=0\)

=> 3x=0 hoặc x+2 = 0

+) 3x = 0 => x =0

+) x+2 = 0 => x = -2