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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
(x + 1)3 - (x + 3)2.(x + 1) + 4x2 + 8 = x3 + 3x2 + 3x + 1 - (x2 + 6x + 9)(x + 1) + 4x2 + 8
= x3 + 7x2 + 3x + 9 - (x3 + 6x2 + 9x + x2 + 6x + 9) = -12x
(x - 2)(x2 + 2x + 4) - (x + 1)3 + 3(x - 1)(x + 1) = x3 + 2x2 + 4x - 2x2 - 4x - 8 - x3 - 3x2 - 3x - 1 + 3(x2 - 1)
= -9 - 3x2 - 3x + 3x2 - 3 = -12 - 3x
(x4 - 52 + 25)(x2 + 5) - (2 + x2)3 + 3(1 + x2)2 = x6 + 5x4 - 8 - 12x2 - 6x4 - x6 + 3(1 + 2x2 + x4) = -x4 - 12x2 - 8 + 3 + 6x2 + 3x4
= 2x4 - 6x2 - 5
1) Có 3 = (22 - 1)
=> BT = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 +1)
= (24 - 1)(24 + 1)(28 + 1)(216 +1)
= (28 - 1)(28 + 1)(216 +1)
= (216 - 1)(216 +1)
= 232 - 1
D=(x2+x+1)(x2-x+1)(x4-x2+1)(x8-x4+1)
\(=\left(\left(x^2+1\right)^2-x^2\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(\left(x^4+1\right)^2-x^4\right)\left(x^8-x^4+1\right).\)
\(=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)=\left(x^8+1\right)-x^8=x^{16}+x^8 +1\)
rút gọn biểu thức
a) (x-2y)2+(x+1)2-(2x+2).(x-2y)
=(x2-4xy+4y2)+(x2+2x+1)-(2x2-4xy+2x-4y)
=x2-4xy+4y2+x2+2x+1-2x2+4xy-2x+4y
=4y2+4y+1