\(1,a^2+3b=b^2+3a\\ \Leftrightarrow a^2-b^2+3b-3a=0\\ \Leftrightarrow\left(a-b\right)\left(a+b-3\right)=0\\ \Leftrightarrow a+b-3=0\left(a\ne b\right)\\ \Leftrightarrow a=3-b\\ \Leftrightarrow\left(3-b\right)^2+3b=10\\ \Leftrightarrow b^2-3b-1=0\\ \Leftrightarrow\left[{}\begin{matrix}b=\dfrac{3+\sqrt{13}}{2}\Rightarrow a=\dfrac{3-\sqrt{13}}{2}\\b=\dfrac{3-\sqrt{13}}{2}\Rightarrow a=\dfrac{3+\sqrt{13}}{2}\end{matrix}\right.\)
Do đó \(a+b\) và \(a^3+b^3\) cùng nhận một giá trị trong cả 2 trường hợp của a,b
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{6}{2}=3\\ab=-1\end{matrix}\right.\Leftrightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=27+3\cdot3=36\)
1.
\(\left\{{}\begin{matrix}a^2+3b=b^2+3a\\a^2+3b=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)-3\left(a-b\right)=0\\a^2+3b=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b-3\right)=0\\a^2+3b=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b-3=0\\a^2+3b=10\end{matrix}\right.\)
\(\Rightarrow a^2+3\left(3-a\right)=10\)
\(\Rightarrow a^2-3a-1=0\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}a+b=3\\ab=-1\end{matrix}\right.\)
\(\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=3^3-3.\left(-1\right).3=36\)
\(2,n\left(n+3\right)\left(n+4\right)\left(n+7\right)+36\\ =\left(n^2+7n\right)\left(n^2+7n+12\right)+36\\ =\left(n^2+7n\right)^2+12\left(n^2+7n\right)+36\\ =\left(n^2+7n+6\right)^2\left(đpcm\right)\)
\(3,P\left(x\right)=x^4+16x^2+64-16x^2=\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\\ 4,P\left(x\right)⋮Q\left(x\right)=x^2-3x+2=\left(x-1\right)\left(x-2\right)\\ \Leftrightarrow\left\{{}\begin{matrix}P\left(1\right)=0\\P\left(2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b-2=0\\2a+b-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)
2.
\(n\left(n+7\right)\left(n+3\right)\left(n+4\right)+36\)
\(=\left(n^2+7n\right)\left(n^2+7n+12\right)+36\)
\(=\left(n^2+7n\right)^2+12\left(n^2+7n\right)+36\)
\(=\left(n^2+7n+6\right)^2\) là SCP
\(5,\\ a,a^3+b^3+3ab\le1\\ \Leftrightarrow a^3+b^3-1+3ab\le0\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\le0\\ \Leftrightarrow\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\le0\\ \Leftrightarrow\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\le0\)
Mà \(a^2+b^2+1+a+b-3ab=\dfrac{1}{2}\left[\left(a+1\right)^2+\left(b+1\right)^2+\left(a-b\right)^2\right]\ge0\)
Do đó \(a+b-1\le0\Leftrightarrow a+b\le1\)
3.
\(x^4+64=\left(x^4+16x^2+64\right)-16x^2=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
4.
Do \(x^2-3x+2\) có 2 nghiệm \(x=\left\{1;2\right\}\) nên P(x) chia hết \(x^2-3x+2\) khi:
\(\left\{{}\begin{matrix}P\left(1\right)=0\\P\left(2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}1-3+a+b=0\\2^3-3.2^2+2a+b=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\2a+b=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)
\(b,1\ge a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{1}{4}\\ \Leftrightarrow E=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15}{16\cdot\dfrac{1}{4}}=\dfrac{1}{2}+\dfrac{60}{16}=\dfrac{17}{4}\\ E_{min}=\dfrac{17}{4}\Leftrightarrow a=b=\dfrac{1}{2}\\ c,F\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}=\dfrac{4}{\left(a+b\right)^2}+\dfrac{1}{2ab}\ge\dfrac{4}{1}+\dfrac{1}{\dfrac{1}{2}}=6\\ F_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)
5.
\(a^3+b^3+3ab\le1\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+3ab-1\le0\)
\(\Leftrightarrow\left(a+b-1\right)\left[\left(a+b\right)^2+a+b+1\right]-3ab\left(a+b-1\right)\le0\)
\(\Leftrightarrow\left(a+b-1\right)\left(a^2-ab+b^2+a+b+1\right)\le0\)
\(\Leftrightarrow\left(a+b-1\right)\left[\left(a-b\right)^2+\left(a+1\right)^2+\left(b+1\right)^2\right]\le0\)
\(\Leftrightarrow a+b-1\le0\)
\(\Leftrightarrow a+b\le1\)
b.
\(E=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{\dfrac{ab}{16ab}}+\dfrac{15}{4\left(a+b\right)^2}\ge\dfrac{2}{4}+\dfrac{15}{4.1}=\dfrac{17}{4}\)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
c.
\(F=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{2}{\left(a+b\right)^2}=\dfrac{6}{\left(a+b\right)^2}\ge\dfrac{6}{1}=6\)
\(6,\Leftrightarrow Ax^2+Ax+5A=3x-1\\ \Leftrightarrow Ax^2+x\left(A-3\right)+5A+1=0\\ \text{PT có nghiệm nên }\Delta=\left(A-3\right)^2-4A\left(5A+1\right)\ge0\\ \Leftrightarrow A^2-6A+9-20A^2-4A\ge0\\ \Leftrightarrow-19A^2-10A+9\ge0\\ \Leftrightarrow-1\le A\le\dfrac{9}{19}\\ \Leftrightarrow A_{min}=-1;A_{max}=\dfrac{9}{19}\)
6.
\(A=\dfrac{3x-1}{x^2+x+5}\Rightarrow Ax^2+Ax+5A=3x-1\)
\(\Leftrightarrow Ax^2+\left(A-3\right)x+5A+1=0\)
\(\Delta=\left(A-3\right)^2-4A\left(5A+1\right)\ge0\)
\(\Rightarrow-19A^2-10A+9\ge0\)
\(\Rightarrow-1\le A\le\dfrac{9}{19}\)
\(\Rightarrow A_{min}=-1\) ; \(A_{max}=\dfrac{9}{19}\)
Do A nguyên \(\Rightarrow\left[{}\begin{matrix}A=-1\\A=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x-1}{x^2+x+5}=-1\\\dfrac{3x-1}{x^2+x+5}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
7.
\(\left(a+b\right)^3-3ab\left(a+b\right)-6ab=-11\)
\(\Leftrightarrow\left(a+b\right)^3+8-3ab\left(a+b+2\right)=-3\)
\(\Leftrightarrow\left(a+b+2\right)\left[\left(a+b\right)^2-2\left(a+b\right)+4\right]-3ab\left(a+b+2\right)=-3\)
\(\Leftrightarrow\left(a+b+2\right)\left[a^2-ab+b^2-2\left(a+b\right)+4\right]=-3\)
\(\Leftrightarrow\left(a+b+2\right)\left[\left(a-b\right)^2+\left(a-2\right)^2+\left(b-2\right)^2\right]=-6< 0\)
\(\Rightarrow a+b+2< 0\)
\(\Rightarrow a+b< -2\)
Lại có:
\(\left(a-b\right)^2+\left(a-2\right)^2+\left(b-2\right)^2\ge\left(a-2\right)^2+\left(b-2\right)^2=a^2+b^2-4\left(a+b\right)+8\)
\(\ge\dfrac{1}{2}\left(a+b\right)^2-4\left(a+b\right)+8\)
Theo cmt: \(a+b< -2\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2>4\\-4\left(a+b\right)>8\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2+\left(a-2\right)^2+\left(b-2\right)^2>\dfrac{1}{2}.4+8+8=18\)
\(\Rightarrow a+b+2=\dfrac{-6}{\left(a-b\right)^2+\left(a-2\right)^2+\left(b-2\right)^2}>\dfrac{-6}{18}=-\dfrac{1}{3}\)
\(\Rightarrow a+b>-2-\dfrac{1}{3}=\dfrac{7}{3}\)