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

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

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

\(=4x+3\)

b) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)

\(=5\left(x^2-4\right)-3+4x+17\)

\(=5x^2-20-3+4x+17\)

\(=5x^2-6+4x\)

a) Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

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

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

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

b) Ta có: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)

\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)

\(=5x^2-20-32x^2+48x-16+17\)

\(=-27x^2+48x-19\)

Ta có : (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 8x + 16 - (x² - 1) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

=> 8x = -1

=> x = \(-\frac{1}{8}\)

Ta có : x² - 4x + 4 =0 

<=> x² - 2.x.2 + 2² = 0

<=> (x - 2)² = 0

=> x - 2 = 0

=> x = 2

cam ơn bạn 

bạn có thể giúp mình tiếp dc ko

b, 5(x + 2) (x - 2 ) - 1/2 (6-8x)² + 17

=5x +10 (x - 2) - 1/2 . 6 - 1/2 . 8x +17

=5x + 10x - 20 - 3 - 4x +17

=15x - 17 -4x + 17 

=15x - 4x -17 + 17

=11x - 0 =11x

a, (x+1)² - (x-1)² - 3(x+1) (x-1)

=(x+1)+(x-1).(x+1)-(x-1) - 3x+3x -3

=2x.0 - 3x

=-3x

a) = x^2 + 2x + 1 - x^2 +2x - 1 -3x^2 +x - x - 1

= - 3x^2 +4x -1

b) =5x^2 + 10x - 10x - 20 - 1/2 .(36 - 96x + 64x^2 ) +17

= 5x^2 - 20 - 18 - 48 x - 32x^2 +17

= -27x^2 - 48x - 3

Chúc bn hok tốt a !

Mình làm thử nha:

a/ \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right).\left(x-1\right)\)

\(=\left(x+1\right)\left(x+1\right)-\left(x-1\right)\left(x-1\right)-\left(3x+3\right).\left(x-1\right)\)

\(=\left[\left(x+1\right)\left(x+1\right)-\left(x-1\right)\left(x-1\right)\right]-4x+\left(-3\right)\)

Từ đó làm tiếp

b/ \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)^4+17\)

\(=\left(5x+10\right)\left(x-2\right)-\left(3-4x\right)^4+17\)

\(=6x+\left(-20\right)-\left(81-256x\right)+17\)

Làm nốt nha