Ta thấy :

S  = cạnh x cạnh

Vậy 64 = 8 x 8

=> Cạnh hình vuông là:8cm

Chu vi là: 8x4=32(cm)

Ta thấy :

S  = cạnh x cạnh

Vậy 64 = 8 x 8

=> Cạnh hình vuông là:8cm

Chu vi là: 8x4=32(cm)

Áp dụng định lý Py-ta-go cho \(\triangle\)DEF vuông tại D ta có :

DE² + DF² = EF²

\(\Rightarrow\) DF² = EF² - DE²

\(\Rightarrow\) DF = \(\sqrt{EF^2-DE^2}\) = \(\sqrt{13^2-5^2}\) = 12 .

Vậy DF = 12 cm 

Xét △DEF có D = 90°

Áp dụng định lí Py-ta-go:

EF² = DE² + DF²

13² = 5² + DF²

DF² = 13² - 5²= 169 - 25= 144

DF= \(\sqrt{144}\)= 12 cm

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

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

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

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