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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: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)
\(=5x^2-20-32x^2+48x-16+17\)
\(=-27x^2+48x-19\)
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) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)
\(=5\left(x^2-4\right)-3+4x+17\)
\(=5x^2-20-3+4x+17\)
\(=5x^2-6+4x\)
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
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
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 !
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
\(5\left(x-2\right)\left(x+2\right)-\frac{1}{2}\left(6-8x\right)^2+17\)
= \(5\left(x^2-4\right)-\frac{1}{2}\left[2\left(3-4x\right)\right]^2+17\)
= \(5x^2-20-2\left(3-4x\right)^2+17\)
= \(5x^2-3-2\left(3-4x\right)^2\)
= \(5x^2-3-2\left(9-24x+16x^2\right)\)
= \(5x^2-3-18+48x-32x^2\)
= \(-27x^2+48x-21\)
= \(-3\left(9x^2-16x+7\right)\)
Rút gọn biểu thức :
a) (x + 1)2 - (x - 1)2 - 3 (x + 1)(x - 1)
b) 5(x + 2)(x -2) - 1/2 (6 - 8x)2 + 17
a, \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1-x+1\right)\left(x+1+x-1\right)-3\left(x^2-1\right)\)
\(=4x-3x^2+3\)
b, \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(36-96x+64x^2\right)+17\)
\(=5x^2-20-18+48x-32x^2+17\)
\(=-27x^2+48x-21\)
chắc hết mờ r chứ bn