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

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

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

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

tĩm,ybiet

B=x^2-2xy+2y^2+2y+1

5( x+2 ) . ( x-2 ) -12 .( 6 - 8x )² +17 

=5.x^2-2^2-12*6^2

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

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

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

\(=9x^2-6\)

chắc hết mờ r chứ bn

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

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

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

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

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

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

\(=5x^2-20+30-32x^2+17=-27x^2+27\)

\(\left(x^2-1^3\right)-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(=x^2-1-\left(x^6-x^4+x^4-x^2+x^2-1\right)\)

\(=x^2-1-x^6+1=x^2-x^6\)

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

= 5( x² - 4 ) - 1/2( 36 - 96x + 64x² ) + 17

= 5x² - 20 - 18 + 48x - 32x² + 17

= -27x² + 48x - 21

2. ( x² - 1³ ) - ( x⁴ + x² + 1 )( x² - 1 )

= ( x² - 1 ) - ( x⁴ + x² + 1 )( x² - 1 )

= ( x² - 1 )[ 1 - ( x⁴ + x² + 1 ) ]

= ( x² - 1 )( 1 - x⁴ - x² - 1 )

= ( x² - 1 )( -x⁴ - x² )

= -x⁶ - x⁴ + x⁴ + x²

= x² - x⁶

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