Hiển nhiên \(c\left(c+1\right)>a\left(a+1\right)\Rightarrow c>a\ge b\)

Nếu \(c\ge2a\Rightarrow c\left(c+1\right)\ge2a\left(2a+1\right)=4a^2+2a\)

Mà \(a\left(a+1\right)+b\left(b-1\right)\le a\left(a+1\right)+a\left(a-1\right)=2a^2\)

\(\Rightarrow2a^2\ge4a^2+2a\Rightarrow2a^2+2a\le0\) (vô lý)

\(\Rightarrow c< 2a\)

Ta có:

\(4a\left(a+1\right)+4b\left(b-1\right)+1=4c\left(c+1\right)+1\)

\(\Leftrightarrow4a\left(a+1\right)+\left(2b-1\right)^2=\left(2c+1\right)^2\)

\(\Leftrightarrow4a\left(a+1\right)=\left(2c+1\right)^2-\left(2b-1\right)^2\)

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

Nếu \(c-b+1\ge a\Rightarrow\left(c-b+1\right)\left(c+b\right)>a\left(a+b\right)>a\left(a+1\right)\) (ktm)

\(\Rightarrow c-b+1< a\) \(\Rightarrow c-b+1\) ko có ước nguyên tố nào là a

\(\Rightarrow c+b⋮a\Rightarrow\dfrac{c+b}{a}\in Z\) (1)

Theo chứng minh ban đầu, ta có \(b\le a< c< 2a\)

\(\Rightarrow a< c+b< 2a+a=3a\Rightarrow1< \dfrac{c+b}{a}< 3\) (2)

(1);(2) \(\Rightarrow\dfrac{c+b}{a}=2\Rightarrow c+b=2a\)

Thế vào (*) \(\Rightarrow a+1=2\left(c-b+1\right)\Rightarrow2c-2b+1=a\)

\(\Rightarrow2\left(2a-b\right)-2b+1=a\Rightarrow3a-4b+1=0\)

\(\Rightarrow3\left(a-1\right)=4\left(b-1\right)\)

\(\Rightarrow b-1⋮3\Rightarrow b-1=3k\Rightarrow b=3k+1\)

\(\Rightarrow a=4k+1\)

\(\Rightarrow c=2a-b=5k+1\)

\(\Rightarrow A=3\left(5k+1\right)-5\left(3k+1\right)=-2\)

a) Có:

 \(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)

câu (b) cho đa thức P (x) = cái gì?

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

=>\(\begin{cases}a^2\le4\\ b^2\le4\\ c^2\le4\end{cases}\)

=>-2<=a<=2 và -2<=b<=2 và -2<=c<=2

=>2-a>=0; 2-b>=0; 2-c>=0

2-a>=0

\(a^2\ge0\)

Do đó: \(a^2(2-a)\ge0\)

=>\(2a^2-a^3\ge0\)

=>\(a^3\le2a^2\)

Chứng minh tương tự, ta cũng có: \(b^3\le2b^2;c^3\le2c^2\)

Do đó: \(a^3+b^3+c^3\le2\left(a^2+b^2+c^2\right)=2\cdot4=8\)

Dấu '=' xảy ra khi \(\begin{cases}a^2\left(2-a\right)=0\\ b^2\left(2-b\right)=0\\ c^2\left(2-c\right)=0\end{cases}\)

=>a;b;c∈{0;2}

mà \(a^2+b^2+c^2=4\)

nên a=2; b=c=0

\(P = a^4 + b^4 + c^4 = 2^4 + 0^4 + 0^4 = 16\)

\(a^3+b^3+c^3-3abc=1\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=1\) (1)

Do \(a^2+b^2+c^2-ab-bc-ca>0\Rightarrow a+b+c>0\)

(1)\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca+\dfrac{1}{a+b+c}\)

\(\Leftrightarrow3a^2+3b^2+3c^2=\left(a+b+c\right)^2+\dfrac{1}{a+b+c}\ge3\)

\(\Rightarrow a^2+b^2+c^2\ge1\)