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2 tháng 9 2016

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

2 tháng 9 2016

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 !

11 tháng 9 2020

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

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

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

11 tháng 9 2020

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

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

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

7 tháng 6 2016

a

(x+1)-(x-1)-3(x+1)(x-1)

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

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

=x+1-x+1-3x+x-1

=x-x-3x+x+1+1-1

=-2x

b,

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

=5x+10(x-2)-1/2(36-64x2)+17

=5x+10x-20-18+32x2+17

=5x+10x-20-18+17+32x2

=15x-21+32x2

7 tháng 6 2016

a

(x+1)-(x-1)-3(x+1)(x-1)

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

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

=x+1-x+1-3x+x-1

=x-x-3x+x+1+1-1

=-2x

b,

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

=5x+10(x-2)-1/2(36-64x2)+17

=5x+10x-20-18+32x2+17

=5x+10x-20-18+17+32x2

=15x-21+32x2

29 tháng 7 2018

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

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

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

\(=9x^2-6\)

23 tháng 1 2020

\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left[\frac{x^2}{x\left(x^2-4\right)}+\frac{-6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{-2}{x-2}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{-1}{x-2}\)