Câu hỏi của Hoàng thị Hiền

Question

1 tính
a) (x+1)(x^2-x+1)-(x-1)(x^2+x+1)
b) x(x-4)(x+4)-(x^2+1)(x^2-1)

Như Khương Nguyễn · Accepted Answer

a,

$\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)$

$=x^3+1-x^3+1=2$

b, $x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)$

$=x\left(x^2-16\right)-\left(x^4-1\right)$

$=x^3-16x-x^4+1$Câu trả lời: a,

$\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)$

$=x^3+1-x^3+1=2$

b, $x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)$

$=x\left(x^2-16\right)-\left(x^4-1\right)$

$=x^3-16x-x^4+1$

Mysterious Person · Answer

a) $\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)$

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

= $x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1$

= $2$

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

= $x\left(x^2-16\right)-\left(x^4-1\right)$ = $x^3-16x-x^4+1$

= $-x^4+x^3-16x+1$Câu trả lời: a) $\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)$

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

= $x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1$

= $2$

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

= $x\left(x^2-16\right)-\left(x^4-1\right)$ = $x^3-16x-x^4+1$

= $-x^4+x^3-16x+1$

truonganhquan · Answer

a,

(x+1)(x2−x+1)−(x−1)(x2+x+1)(x+1)(x2−x+1)−(x−1)(x2+x+1)

=x3+1−x3+1=2=x3+1−x3+1=2

b, x(x−4)(x+4)−(x2+1)(x2−1)x(x−4)(x+4)−(x2+1)(x2−1)

=x(x2−16)−(x4−1)=x(x2−16)−(x4−1)

=x3−16x−x4+1Câu trả lời: a,

(x+1)(x2−x+1)−(x−1)(x2+x+1)(x+1)(x2−x+1)−(x−1)(x2+x+1)

=x3+1−x3+1=2=x3+1−x3+1=2

b, x(x−4)(x+4)−(x2+1)(x2−1)x(x−4)(x+4)−(x2+1)(x2−1)

=x(x2−16)−(x4−1)=x(x2−16)−(x4−1)

=x3−16x−x4+1

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