Theo đề ra, ta có:

\(a+b+c=1\)

\(\Rightarrow a=1-b-c\)

\(\Rightarrow a+1=\left(1-b\right)+\left(1-c\right)\)

Theo AM-GM, ta có:

\(a+1=\left(1-b\right)+\left(1-c\right)\ge2\sqrt{\left(1-b\right)\left(1-c\right)}\) (*)

Chứng minh tương tự:

\(b+1\ge2\sqrt{\left(1-a\right)\left(1-c\right)}\) (**)

\(c+1\ge2\sqrt{\left(1-a\right)\left(1-b\right)}\) (***)

Từ (*)(**)(***) \(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{\left[\left(1-a\right)\left(1-b\right)\left(1-c\right)\right]^2}=8\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Ba số đó là: 2;3;4

Ta có

\(\left(a^4+b^4+c^4\right)\left(a^3+b^3+c^3\right)=\)

\(=a^7+a^4b^3+a^4c^3+a^3b^4+b^7+b^4c^3+a^3c^4+b^3c^4+c^7\)

\(\Rightarrow\left(a^7+b^7+c^7\right)=\left(a^4+b^4+c^4\right)\left(a^3+b^3+c^3\right)-a^3b^3\left(a+b\right)-b^3c^3\left(b+c\right)-a^3c^3\left(a+c\right)=\)

Do a+b+c=0

\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)

\(=\left(a^7+b^7+c^7\right)=\left(a^4+b^4+c^4\right)\left(a^3+b^3+c^3\right)+a^3b^3c+b^3c^3a+a^3c^3b=\)

\(=\left(a^4+b^4+c^4\right)\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]+abc\left(a^2b^2+b^2c^2+a^2c^2\right)=\)

\(=\left(a^4+b^4+c^4\right).3abc+abc\left(a^2b^2+b^2c^2+a^2c^2\right)=\)

\(=abc.\left[3\left(a^4+b^4+c^4\right)+a^2b^2+b^2c^2+a^2c^2\right]\) (1)

Mà

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=\)

\(=\left[\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\right]^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=\)

\(=4\left(ab+bc+ca\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=\)

\(=4\left[\left(a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc\right)\right]-2\left(a^2b^2+b^2c^2+a^2c^2\right)=\)

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

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

\(\Rightarrow a^2b^2+b^2c^2+a^2b^2=\dfrac{a^4+b^4+c^4}{2}\) Thay vào (1) ta có

\(a^7+b^7+c^7=abc.\left[3\left(a^4+b^4+c^4\right)+\dfrac{a^4+b^4+c^4}{2}\right]=\)

\(=7.abc.\dfrac{a^4+b^4+c^4}{2}\)

\(\Rightarrow2\left(a^7+b^7+c^7\right)=7.abc.\left(a^4+b^4+c^4\right)\left(đpcm\right)\)

e) \(149,47-108,7=40,77\)

     \(9,204\) x \(8,2=75,4728\)

g)  \(6750-39,72=6710,28\)

      \(0,504\) x \(72=36,288\)

h)   \(123,56+374,8=498,36\)

      \(112,56:28=4,02\)

i)    \(962:58=16,5862069\)

     \(78,24:2,4=32,6\)

Ta có

\(\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)=a^5+a^2b^3+a^2c^3+a^3b^2+b^5+b^2c^3+a^3c^2+b^3c^2+c^5\)

\(\Rightarrow a^5+b^5+c^5=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)-a^2b^2\left(a+b\right)-b^2c^2\left(b+c\right)-a^2c^2\left(a+c\right)\)

Do a+b+c=0

=> a+b=-c; b+c=-a; a+c=-b

\(\Rightarrow a^5+b^5+c^5=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)+a^2b^2c+ab^2c^2+a^2bc^2=\)

\(=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)+abc\left(ab+bc+ac\right)=\)

\(=\left(a^2+b^2+c^2\right)\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]+abc\left(ab+bc+ac\right)=\)

\(=\left(a^2+b^2+c^2\right).\left[\left(-c^3\right)-3ab.\left(-c\right)+c^3\right]+abc\left(ab+bc+ac\right)=\)

\(=\left(a^2+b^2+c^2\right).3abc+abc\left(ab+bc+ab\right)=\)

\(=abc.\left[3\left(a^2+b^2+c^2\right)+ab+bc+ac\right]=\)

\(=abc\left[\dfrac{5}{2}.\left(a^2+b^2+c^2\right)+\dfrac{a^2+b^2+c^2+2ab+2bc+2ac}{2}\right]=\)

\(=abc.\left[\dfrac{5}{2}.\left(a^2+b^2+c^2\right)+\dfrac{\left(a+b+c\right)^2}{2}\right]=\)

\(=abc.\dfrac{5}{2}.\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\dfrac{a^5+b^5+c^5}{5}=abc.\dfrac{a^2+b^2+c^2}{2}\left(đpcm\right)\)