Ta có (a + b + c)² = a² + b² + c²

=> a² + b² + c² + 2ab + 2bc + 2ca = a² + b² + c²

=> 2(ab + bc + ca) = 0

=> ab + bc + ca = 0

=> \(\frac{abc}{c}+\frac{abc}{a}+\frac{abc}{b}=0\)

=> \(abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\left(\text{Vì }abc\ne0\right)\)

=> \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{a^2b}+\frac{3}{ab^2}=-\frac{1}{c^3}\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)(Vì 1/a + 1/b = -1/c)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)(đpcm)

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

a, Thay x = -3 ta được : 

\(=\left(-3+1\right)^{-3}-2\left[\left(-3\right)^2-3-2\right]+2\)

\(=-\frac{1}{8}-8+2=-\frac{1}{8}-\frac{64}{8}+\frac{16}{8}=\frac{-49}{8}\)

b, Ta có : \(m=0\)hay \(\left(x+1\right)^x-2\left(x^2+x-2\right)+2=0\)

... =))? 

a) x³ = 25x

=> x³ - 25x = 0

=> x(x² - 25) = 0

=> x(x - 5)(x + 5) = 0

=> x = 0 hoặc x - 5 = 0 hoặc x + 5 = 0

=> x = 0 hoặc x = 5 hoặc x = -5

b) x² - 6x + 8 = 0

=> x² - 6x + 9 - 1 = 0

=> (x - 3)² - 1² = 0

=> (x - 3 - 1)(x - 3 + 1) = 0

=> (x - 4)(x - 2) = 0

=> \(\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

a. 2x(x + y) - y(y + 2x) = 2x² + 2xy - y² - 2xy = 2x² - y²

b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

Phần c nản quá.