\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

\(=4294967295\)

_Chúc bạn học tốt!_

Bạn ơi, có lẽ đề bài mình cần sửa 1 chút.
Mình sửa lại thành 125x^3 - 10x^2 + 2x -1 nhé.
Giải: \(125x^3-10x^2+2x-1\)

= \(125x^3-25x^2+15x^2-3x+5x-1\)

= \(25x^2\left(5x-1\right)+3x\left(5x-1\right)+\left(5x-1\right)\)

=\(\left(5x-1\right)\left(25x^2+3x+1\right)\)

_Chúc bạn học tốt!_

Ta có: a+b+c=0
=> (a+b+c)^2=0
=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 0
=> 2ab + 2bc + 2ac = -1 ( do a^2 + b^2 + c^2 = 1 )
=> ( 2ab + 2bc + 2ac )^2 = (-1)^2
=> \(4a^2b^2+4b^2c^2+4a^2c^2+8ab^2c+8abc^2+8a^2bc=1\)

=>\(4a^2b^2+4b^2c^2+4a^2c^2+8abc\left(a+b+c\right)=1\)

=>2\(\left(2a^2b^2+2b^2c^2+2a^2c^2\right)=1\)

(do a+b+c=0)
=>\(\left(2a^2b^2+2b^2c^2+2a^2c^2\right)=\dfrac{1}{2}\)

Lại có: a^2 + b^2 + c^2 =1
=> (a^2 + b^2 + c^2 )^2 = 1
=> a^4 + b^4 + c^4 + \(\left(2a^2b^2+2b^2c^2+2a^2c^2\right)=1\)

=> a^4 + b^4 + c^4 + 1/2 = 1
=> a^4 + b^4 + c^4 = 1/2
Mình làm hơi dài thông cảm! Có gì khó hiểu hỏi mình .

Ta có:
\\(x:y:z=\\dfrac{2}{5}:\\dfrac{3}{4}:\\dfrac{1}{3}\\)

\\(\\Rightarrow\\dfrac{x}{\\dfrac{2}{5}}=\\dfrac{y}{\\dfrac{3}{4}}=\\dfrac{z}{\\dfrac{1}{3}}\\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\\(\\dfrac{x}{\\dfrac{2}{5}}=\\dfrac{y}{\\dfrac{3}{4}}=\\dfrac{z}{\\dfrac{1}{3}}=\\dfrac{x-z}{\\dfrac{2}{5}-\\dfrac{1}{3}}=\\dfrac{-4,8}{\\dfrac{1}{15}}=-72\\\\ \\Rightarrow x=\\left(-72\\right).\\dfrac{2}{5}=-28,8\\\\ y=\\left(-72\\right).\\dfrac{3}{4}=-54\\\\ z=\\left(-72\\right).\\dfrac{1}{3}=-24\\\\ \\Rightarrow x=-28,8;y=-54;z=-24\\)