1. b3+b= 3                                       

(b3+b)=3                            

b.(3+1)=3

b. 4= 3

b=\(\dfrac{3}{4}\)

a3+a= 3                                       b3

(a3+a)=3                            

a.(3+1)=3

a. 4= 3

a=\(\dfrac{3}{4}\)

2

1.

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)

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

Ta có:

\(\dfrac{\left(a+2b\right)^2+\left(b+2c\right)^2+\left(c+2a\right)^2}{\left(a-2b\right)^2+\left(b-2c\right)^2+\left(c-2a\right)^2}\)

\(=\dfrac{a^2+4b^2+4ab+b^2+4c^2+4bc+c^2+4a^2+4ca}{a^2+4b^2-4ab+b^2+4c^2-4bc+c^2+4a^2-4ca}\)

\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)

\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)

\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)

b.

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3abc\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)

Mk ms tìm được GTNN thôi!

Ta có: A = a³ + b³ = (a + b)(a² + b² - ab) = (a + b)(1 - ab)

Áp dụng BĐT Cô-si cho 2 số ko âm a² và b² ta có:

a² + b² \(\ge\) 2ab

\(\Leftrightarrow\) 1 \(\ge\) 2ab

\(\Leftrightarrow\) 1 - 2ab \(\ge\) 0

\(\Leftrightarrow\) 1 - ab \(\ge\) ab

\(\Rightarrow\) A \(\ge\) ab(a + b)

Dấu "=" xảy ra khi và chỉ khi a = b = \(\sqrt{0,5}\)

\(\Rightarrow\) A \(\ge\) 0,5 . 2\(\sqrt{0,5}\) = \(\sqrt{0,5}\)

Vậy ...

Chúc bn học tốt!

\(a^2+b^2=1\Rightarrow\left\{{}\begin{matrix}0\le a\le1\\0\le b\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3\le a^2\\b^3\le b^2\end{matrix}\right.\)

\(\Rightarrow a^3+b^3\le a^2+b^2=1\)

\(A_{max}=1\) khi \(\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)

\(a^3+a^3+\left(\dfrac{1}{\sqrt{2}}\right)^3\ge\dfrac{3}{\sqrt{2}}a^2\)

\(b^3+b^3+\left(\dfrac{1}{\sqrt{2}}\right)^3\ge\dfrac{3}{\sqrt{2}}b^2\)

Cộng vế:

\(2\left(a^3+b^3\right)+\dfrac{\sqrt{2}}{2}\ge\dfrac{3}{\sqrt{2}}\left(a^2+b^2\right)=\dfrac{3\sqrt{2}}{2}\)

\(\Rightarrow a^3+b^3\ge\dfrac{\sqrt{2}}{2}\)

\(A_{min}=\dfrac{\sqrt{2}}{2}\) khi \(a=b=\dfrac{\sqrt{2}}{2}\)

Ta có: A = a³ + b³ = (a + b)(a² + b² - ab) = (a + b)(1 - ab)

Áp dụng BĐT Cô-si cho 2 số (a + b)² và 1 ko âm ta có:

\(\dfrac{\left(a+b\right)^2+1}{2}\ge a+b\)

\(\Leftrightarrow\) \(\dfrac{a^2+b^2+2ab+1}{2}\ge a+b\)

\(\Leftrightarrow\) \(\dfrac{2+2ab}{2}\ge a+b\)

\(\Leftrightarrow\) 1 + ab \(\ge\) a + b

\(\Leftrightarrow\) (1 - ab)(1 + ab) \(\ge\) A

\(\Leftrightarrow\) 1 - a²b² \(\ge\) A

Dấu "=" xảy ra \(\Leftrightarrow\) ab = 1; a² + b² = 1

Khi đó: A \(\le\) 0

Vậy ...

Chúc bn học tốt!

Bài 3: 

a: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=7^2-4\cdot12=1\)

b: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=7^3-3\cdot12\cdot7\)

\(=343-252=91\)

\(a^2+b^2=a^3+b^3=a^4+b^4\)

\(\Rightarrow\left(a^3+b^3\right)^2=\left(a^2+b^2\right)\left(a^4+b^4\right)\)

\(\Rightarrow a^6+b^6+2a^3b^3=a^6+b^6+a^2b^4+a^4b^2\)

\(\Rightarrow2a^3b^3=a^2b^2\left(a^2+b^2\right)\)

\(\Rightarrow2ab=a^2+b^2\)

\(\Rightarrow\left(a-b\right)^2=0\)

\(\Rightarrow a=b\)

Thế vào \(a^2+b^2=a^3+b^3\)

\(\Rightarrow a^2+a^2=a^3+a^3\Rightarrow2a^3=2a^2\Rightarrow a=b=1\)

\(\Rightarrow a+b=2\)

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)