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Đầu tiên ta có nhận xét là y>0
Ta có:\(y=\frac{3x^2-x+1}{2x^2+x+1}\)
Ta có:\(6x^2+3x+3-3x^2+x-1=3x^2+4x+2=2\left(x+1\right)^2+x^2>0\)
\(\Rightarrow3x^2-x+1< 3\left(2x^2+x+1\right)\)
\(\Rightarrow y< 3\)
\(\Rightarrow y\in\left\{1;2\right\}\)
\(\Rightarrow x=\left\{0;2;\frac{-3+\sqrt{5}}{2};\frac{-3-\sqrt{5}}{2}\right\}\)
Ta có:\(S=2a+\frac{1}{a^2}\)
\(A=8a+8a+\frac{1}{a^2}-14a\)
\(A\ge3\sqrt[3]{8a\cdot8a\cdot\frac{1}{a^2}}-14\cdot\frac{1}{2}\)
\(A\ge14-7=5\)
"="<=>a=1/2
\(\left(x^2-6x\right)^2-2\left(x-3\right)^2=81\)
\(\Leftrightarrow\left(x^2-6x\right)^2-81=2\left(x-3\right)^2\)
\(\Leftrightarrow\left(x^2-6x-9\right)\left(x^2-6x+9\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow\left(x^2-6x-9\right)\left(x-3\right)^2=2\left(x-3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2-6x-9=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=3+2\sqrt{5}\\x=3-2\sqrt{5}\end{matrix}\right.\)
\(A=2x^2-x+1\)
\(A=\left(2x^2-x+\frac{1}{8}\right)+\frac{7}{8}\)
\(A=\left(\sqrt{2}x-\frac{1}{2\sqrt{2}}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)
"="<=>\(x=\frac{1}{4}\)
Ta có:\(\frac{a^3}{a^2+ab+b^2}=\frac{a\left(a^2+ab+b^2\right)-ab\left(a+b\right)}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\)
Lại có:\(a^2+ab+b^2\ge3ab\)
\(\Rightarrow a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3ab}=a-\frac{a+b}{3}\)
\(\Rightarrow\sum\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
"="<=>a=b=c
a)\(5x^2+13y^2+6xy=12x-4y\)
\(\Leftrightarrow5x^2+6x\left(y-2\right)+13y^2+4y=0\)
pt có nghiệm:\(\Delta'=9\left(y-2\right)^2-65y^2-20y\ge0\)
\(\Leftrightarrow9y^2-36y+36-65y^2-20y\ge0\)
\(\Leftrightarrow-56y^2-56y+36\ge0\)
Mà \(y\in Z\)\(\Rightarrow-1\le y\le0\)
\(\Rightarrow y=0;1\)
Thay vào tìm x
\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a^2+b^2+c^2}\)
\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{9-2\left(ab+bc+ca\right)}\)
\(P=\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ca}+\frac{1}{9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(P\ge\frac{16}{3ab+3bc+3ca+9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{9}{ab+bc+ca}\right)\)
\(P\ge\frac{16}{9+ab+bc+ca}+\frac{6}{ab+bc+ca}\)
Sử dụng đánh giá quen thuộc:\(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Rightarrow ab+bc+ca\le3\)
\(\Rightarrow P\ge\frac{16}{9+3}+\frac{6}{3}=2+\frac{4}{3}=\frac{10}{3}\)
"="<=>a=b=c=1