Tui lớp 8 nhg tui nhớ ko nhầm thì sách có giải mà!

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

= \(x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\)

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

f) \(x^3-6x^2-x+30=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)

= \(x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

= \(\left(x-5\right)\left(x^2-x-6\right)\)

= \(\left(x-5\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]\)

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

c) \(x^3-9x^2+6x+16=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)

= \(x^2\left(x-8\right)+x\left(x-8\right)-2\left(x-8\right)\)

= \(\left(x-8\right)\left(x^2-x-2\right)\)

= \(\left(x-8\right)\left(x^2+x-2x-2\right)\)

= \(\left(x-8\right)\left[x\left(x+1\right)-2\left(x+1\right)\right]\)

= \(\left(x-8\right)\left(x+1\right)\left(x-2\right)\)

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

= \(x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)

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

a) \(x^3-7x+6=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)

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

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

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

= \(\left(x-2\right)\left[x\left(x-1\right)+3\left(x-1\right)\right]\)

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

b) Sửa cái đề nha!

\(x^3+5x^2+8x+4\) = \(\left(x^2+2x^2\right)+\left(3x^2+6x\right)+\left(2x+4\right)\)

= \(x^2\left(x+2\right)+3x\left(x+2\right)+2\left(x+2\right)\)

= \(\left(x+2\right)\left(x^2+x+2x+2\right)\)

= \(\left(x+2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

= \(\left(x+2\right)^2\left(x+1\right)\)

Làm tắt thôi ha!

Gọi CD là a (cm); CR là b (cm) ( a >4; b > 0)

=> S của hcn là ab ( cm²)

Theo bài ra ta có HPT:

\(\left\{{}\begin{matrix}\left(a+5\right)\left(b+5\right)=ab+200\\\left(a-4\right)\left(b+1\right)=ab-44\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}5\left(a+b+5\right)=200\\a-4b-4=-44\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a+b=35\\a-4b=-40\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=20\\b=15\end{matrix}\right.\) (TM)

Vậy.................................

Sửa đề: A= x¹⁴- 10x¹³ + 10x¹² - 10x¹¹+....+ 10x² - 10x + 10

Thay x = 9 vào A ta được:

A = \(9^{14}-\left(9+1\right).9^{13}+\left(9+1\right)9^{12}-....+9+1\)

=> A = \(9^{14}-9^{14}-9^{13}+9^{12}-9^{12}+....-9+9+1\)

=> A = 1

Câu c tui nhầm đề! Sửa lại:

\(x^3+2x-3=x^3-x^2+x^2-x+3x-3\)

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

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

Gợi ý:

Tích đó = 0 <=> 1 trong hai thừa số = 0

Th1: x^2 = 0 thì x = mấy?

Th2: 2/3 - 5x = 0 thì 5x = 2/3

Vậy x = mấy?