a. Đề bài em ghi sai thì phải

Vì:

\(x+y=2\left(\sqrt{x-3}+\sqrt{y-3}\right)\)

\(\Leftrightarrow\left(x-3-2\sqrt{x-3}+1\right)+\left(y-3-2\sqrt{y-3}+1\right)+4=0\)

\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+4=0\) (vô lý)

b.

Xét hàm \(f\left(x\right)=x^3+ax^2+bx+c\)

Hàm đã cho là hàm đa thức nên liên tục trên mọi khoảng trên R

Hàm bậc 3 nên có tối đa 3 nghiệm

\(f\left(-2\right)=-8+4a-2b+c>0\)

\(f\left(2\right)=8+4a+2b+c< 0\)

\(\Rightarrow f\left(-2\right).f\left(2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc (-2;2)

\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=x^3\left(1+\dfrac{a}{x}+\dfrac{b}{x^2}+\dfrac{c}{x^3}\right)=+\infty.\left(1+0+0+0\right)=+\infty\)

\(\Rightarrow\) Luôn tồn tại 1 số thực dương n đủ lớn sao cho \(f\left(n\right)>0\)

\(\Rightarrow f\left(2\right).f\left(n\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(2;n\right)\) hay \(\left(2;+\infty\right)\)

Tương tự \(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=-\infty\Rightarrow f\left(-2\right).f\left(m\right)< 0\Rightarrow f\left(x\right)\) luôn  có ít nhất 1 nghiệm thuộc \(\left(-\infty;-2\right)\)

\(\Rightarrow f\left(x\right)\) có đúng 3 nghiệm pb \(\Rightarrow\) hàm cắt Ox tại 3 điểm pb

Ta có : \(a^3+b^3+c^3=3abc\)

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

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

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

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

Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow\left\{{}\begin{matrix}a=b=c\\a;b;c\ne0\end{matrix}\right.\) ( nhấn 2 lên rồi nhóm cặp )

Lại có : \(Q=\frac{a^2+3b^2+5c^2}{\left(a+b+c\right)^2}=\frac{9a^2}{9a^2}=1\)

=> ĐPCM

Lời giải:

Có: \(\left\{\begin{matrix} a+b+c=9\\ a^2+b^2+c^2=27\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (a+b+c)^2=81\\ a^2+b^2+c^2=27\end{matrix}\right.\)

\(\Rightarrow (a+b+c)^2-(a^2+b^2+c^2)=54\)

\(\Leftrightarrow 2(ab+bc+ac)=54\Leftrightarrow ab+bc+ac=27\)

Do đó: \(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow \frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}=0(*)\)

Ta thấy: \((a-b)^2; (b-c)^2; (c-a)^2\geq 0\forall a,b,c\in\mathbb{R}\)

Suy ra \((*)\) xảy ra khi và chỉ khi

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

Khi đó: \(a=b=c=\frac{9}{3}=3\) (thỏa mãn)

\(P=(a-2)^{2015}+(b-3)^{2016}+(c-4)^{2017}=1^{2015}+0^{2016}+(-1)^{2017}\)

\(P=1+0+(-1)=0\)

gt \(\Rightarrow\left\{{}\begin{matrix}b\left(a^2+2ac+c^2\right)+ac\left(a+c\right)+b^2\left(a+c\right)=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\\a^{2013}+b^{2013}+c^{2013}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=0\Rightarrow a^{2013}+b^{2013}=0\\b+c=0\Rightarrow b^{2013}+c^{2013}=0\\a+c=0\Rightarrow a^{2013}+c^{2013}=0\end{matrix}\right.\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow Q=1\)

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

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

Nếu \(a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Rightarrow a=b=c\)

Khi đó \(A=2^3=8\)

Nếu \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

Thay vào ta được:

\(A=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-abc}{abc}=-1\)

Vậy A = 8 hoặc A = -1

Ta có:

\(a^3+b^3+c^3=3abc\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Do a+b+c khác ) nên:

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

\(\Leftrightarrow\frac{1}{2}[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2]=0\)

\(\Rightarrow a=b=c\)

Do đó:

Q=\(\frac{a^2+3b^2+5c^2}{\left(a+b+c\right)^2}=\frac{9a^2}{9a^2}=1\)

có giá trị ko đổi

ĐKXĐ: \(abc\ne0\)

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

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

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

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

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

TH1: \(a+b+c=0\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)

TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)