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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: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)
\(=5x^2-20-32x^2+48x-16+17\)
\(=-27x^2+48x-19\)
5( x+2 ) . ( x-2 ) -12 .( 6 - 8x )2 +17
=5.x^2-2^2-12*6^2
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) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)
\(=5\left(x^2-4\right)-3+4x+17\)
\(=5x^2-20-3+4x+17\)
\(=5x^2-6+4x\)
\(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\frac{1}{2}\left(36-96x+64x^2\right)+17\)
\(=5x^2-20+30-32x^2+17=-27x^2+27\)
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
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
\(5\left(x-2\right)\left(x+2\right)-\frac{1}{2}\left(6-8x\right)^2+17\)
= \(5\left(x^2-4\right)-\frac{1}{2}\left[2\left(3-4x\right)\right]^2+17\)
= \(5x^2-20-2\left(3-4x\right)^2+17\)
= \(5x^2-3-2\left(3-4x\right)^2\)
= \(5x^2-3-2\left(9-24x+16x^2\right)\)
= \(5x^2-3-18+48x-32x^2\)
= \(-27x^2+48x-21\)
= \(-3\left(9x^2-16x+7\right)\)