\(\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}-a-b-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ca-bc-ab}{c+a}+\dfrac{c^2+ab-ca-bc}{a+b}\ge0\)

\(\Leftrightarrow\left(a^2+bc-ab-ac\right)\left(a+b\right)\left(c+a\right)+\left(b^2+ca-bc-ab\right)\left(b+c\right)\left(a+b\right)+\left(c^2+ab-ca-bc\right)\left(b+c\right)\left(c+a\right)\ge0\Leftrightarrow a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2\ge0\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)\ge0\left(đúng\right)\)

\(dấu"="\Leftrightarrow a=b=c\)

\(N1=\dfrac{m^2+2m-8}{m^2-2m+1}\) \(do:m\ne1\Rightarrow m^2-2m+1=\left(m-1\right)^2>0\)

\(\Rightarrow N\left(m^2-2m+1\right)=m^2+2m-8\Leftrightarrow Nm^2-2mN+N=m^2+2m-8\)

\(\Leftrightarrow\left(N-1\right)m^2-2m\left(N+1\right)+N+8=0\)

\(TH:N=1\Rightarrow m=2,25\left(tm\right)\)

\(TH:N\ne1\Rightarrow\Delta'\ge0\Leftrightarrow\left(N+1\right)^2-\left(N+8\right)\left(N-1\right)\ge0\Leftrightarrow N\le\dfrac{9}{5}\Leftrightarrow m=3,5\)

\(từ:2TH\Rightarrow maxN=\dfrac{9}{5}\Leftrightarrow m=3,5\)

(mấy ý kia làm tương tự)

\(\left(dk:x\ge0\right)C=2x+3\sqrt{x}-28=2\left(\sqrt{x}+\dfrac{3}{4}\right)^2-\dfrac{233}{8}\ge2\left(\dfrac{3}{4}\right)^2-\dfrac{233}{8}=-28\Rightarrow dấu"="\Leftrightarrow x=0\)

\(d;E=\sqrt{\left(x-2011\right)^2+\left(x-1\right)^2}\Rightarrow\sqrt{2}.E=\sqrt{\left[\left(2011-x\right)^2+\left(x-1\right)^2\right]\left(1+1\right)}\ge2011-1=2010\Rightarrow E\ge\dfrac{2010}{\sqrt{2}}\)

\(dấu"="\Leftrightarrow2011-x=x-1\Leftrightarrow x=1006\)

\(F=\sqrt{x-2\sqrt{x}-3}\ge0\Rightarrow dấu"="\Leftrightarrow x-2\sqrt{x}-3=0\Leftrightarrow x=9\)

có thể làm cách khác không cần nhân căn 3

\(M=a\sqrt{b\left(a+2b\right)}+b\sqrt{a\left(b+2a\right)}\le\sqrt{\left(a^2+b^2\right)\left[b\left(a+2b\right)+a\left(b+2a\right)\right]}\left(bunhia\right)\)

\(=\sqrt{\left(a^2+b^2\right)\left[2ab+2\left(a^2+b^2\right)\right]}\le\sqrt{2\left[2.\dfrac{a^2+b^2}{2}+2.2\right]}\le\sqrt{2\left(2+2.2\right)}=2\sqrt{3}\)

\(dấu"="\Leftrightarrow a=b=1\)

\(1\left(đk:x\le2\right).A=x+\sqrt{2-x}\)

\(\Leftrightarrow A-x=\sqrt{2-x}\Leftrightarrow A^2-2Ax+x^2=2-x\)

\(\Leftrightarrow x^2-x\left(2A-1\right)+A^2-2=0\)

\(\Rightarrow\Delta\ge0\Leftrightarrow\left(2A-1\right)^2-4\left(A^2-2\right)\ge0\Leftrightarrow A\le\dfrac{9}{4}\)

\(dấu"="\Leftrightarrow x=\dfrac{7}{4}\)

\(3\left(đk:-1\le x\le1\right).A=\sqrt{1-x}+\sqrt{1+x}\le\sqrt{2\left(1-x+1+x\right)}=2\)

\(A^2=2+2\sqrt{\left(1-x\right)\left(1+x\right)}\ge2\Rightarrow A\ge\sqrt{2}\)

\(\Rightarrow max=2\Leftrightarrow x=0;min=\sqrt{2}\Leftrightarrow x=-1\)

\(2.A=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\ge\left|2\right|=2\)

\(4.A=\sqrt{x^2+1}+\sqrt{\left(x-1\right)^2+4}=\sqrt{x^2+1}+\sqrt{\left(1-x\right)^2+2^2}\ge\sqrt{\left(x+1-x\right)^2+\left(1+2\right)^2}=\sqrt{10}\left(minicopxki\right)\)

\(dấu"="\Leftrightarrow\dfrac{x}{1}=\dfrac{1-x}{2}\Leftrightarrow x=\dfrac{1}{3}\)

đó là nhân điểm rơi cosi \(x+y\ge2\sqrt{xy}\) đảm bảo dấu"=" \(\Leftrightarrow x=y\)

\(bài\) \(có:a=b=1\) \(thay\) \(vào:\) \(\sqrt{b\left(a+2b\right)}\) \(=\sqrt{1.\left(1+2\right)}=\sqrt{1.3}\)

vậy dùng cosi luôn thì dấu"=" tại 1=3(vô) lí nên làm phải nhân với \(\sqrt{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(1-\dfrac{12}{y+3x}\right)^2x=4\\\left(1+\dfrac{12}{y+3x}\right)^2y=36\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(1-\dfrac{12}{y+3x}\right)^2=\dfrac{4}{x}\left(1\right)\\\left(1+\dfrac{12}{y+3x}\right)^2=\dfrac{36}{y}\left(2\right)\end{matrix}\right.\)

\(\)\((do:x;y=0\) \(không\) \(là\) \(nghiệm\) \(của\) \(hpt\)\()\)\(\left(dk:x;y\ge0\right)\)

\(\Rightarrow\left(1+\dfrac{12}{y+3x}+1-\dfrac{12}{y+3x}\right)\left(1+\dfrac{12}{y+3x}-1+\dfrac{12}{y+3x}\right)=\dfrac{36}{y}-\dfrac{4}{x}\)

\(\Leftrightarrow\dfrac{48}{y+3x}=\dfrac{36}{y}-\dfrac{4}{x}=\dfrac{36x-4y}{xy}\)

\(\Leftrightarrow48xy=\left(36x-4y\right)\left(y+3x\right)\)

\(\Leftrightarrow48xy=36xy+108x^2-4y^2-12xy\Leftrightarrow108x^2-24xy-4y^2=0\)

\(\Leftrightarrow108\left(\dfrac{x}{y}\right)^2-24\left(\dfrac{x}{y}\right)-4=0\)

\(đặt:\dfrac{x}{y}=a\Rightarrow108a^2-24a-4=0\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{1}{3}\Rightarrow y=3x\\a=-\dfrac{1}{9}\Rightarrow y=-9x\end{matrix}\right.\)

\(với:y=3x\Rightarrow\)\(\left(2\right)\Leftrightarrow\left(1+\dfrac{12}{6x}\right)^2=\dfrac{36}{3x}\)

đến đây đơn giản rồi giải bình thường\(\Rightarrow x=2\left(2+\sqrt{3}\right)\Rightarrow y=3x=6\left(2+\sqrt{3}\right)\)(thỏa)

\(với:y=-9x\Rightarrow\left(2\right)\Leftrightarrow\left(1+\dfrac{12}{-6x}\right)^2=\dfrac{36}{-9x}\)

giải pt =>vô nghiệm

\(\Rightarrow S=\left\{2\left(2+\sqrt{3}\right);6\left(2+\sqrt{3}\right)\right\}\)

\(\left\{{}\begin{matrix}x^3-y^3+3y^2-3x=2\left(1\right)\\x^2+\sqrt{1-x^2}-3\sqrt{2-y-y^2}=-2\left(2\right)\end{matrix}\right.\)

\(đk:\left\{{}\begin{matrix}1-x^2\ge0\\2-y-y^2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\-2\le y\le1\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^3-y^3+3y^2-3x-2=0\)

\(\Leftrightarrow\left(x-y+1\right)\left(x^2+xy-x+y^2-2y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y+1=0\Leftrightarrow x=y-1\\x^2+xy-x+y^2-2y-2=0\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow\left(y-1\right)^2+\sqrt{-y^2+2y-1}-3\sqrt{2-y-y^2}=-2\left(4\right)\)

bạn xem lại đề xem có ghi nhầm chỗ nào không vì nghiệm xấu quá

\(b;\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2+3=0\left(1\right)\\\sqrt[4]{x}+\sqrt{32-x}+6y-24=0\left(2\right)\end{matrix}\right.\)

\(đặt:\left\{{}\begin{matrix}\sqrt[4]{32-x}=a\ge0\\\sqrt[4]{x}=b\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{32-x}=a^2\\\sqrt{x}=b^2\end{matrix}\right.\)

\(\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}b^2+a-y^2+3=0\left(1\right)\\b+a^2+6y-24=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)+\left(2\right)\Leftrightarrow a^2+b^2+a+b-y^2+6y-21=0\Leftrightarrow a^2+b^2+a+b=y^2-6y+21\)

\(có:a^2+b^2\le\sqrt{2\left(a^4+b^4\right)}=\sqrt{2.32}=8\left(bunhia\right)\)

\(có:a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2.8}=4\)

\(\Rightarrow a^2+b^2+a+b\le12\)

\(mà:y^2-6y+21=\left(y-3\right)^2+12\ge12\)

\(\Rightarrow dấu"="xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}a=b\\y=3\end{matrix}\right.\)

\(\Rightarrow a^4=b^4\Leftrightarrow x=32-x\Leftrightarrow x=16\)

\(x=16;y=3\) \(thử\) \(lại\) \(thấy\) \(thỏa\)

\(đặt:\sqrt{x+y+1}=a\ge0;\sqrt{3x+3y}=b\ge0\)\(\left(đk:x+y+1\ge0;3x+3y\ge0\right)\)

\(\Rightarrow b^2-a^2=3\left(x+y\right)-\left(x+y+1\right)=2\left(x+y\right)-1\Leftrightarrow\left(x+y\right)=\dfrac{b^2-a^2+1}{2}\)

\(\Rightarrow a+1=4\left(\dfrac{b^2-a^2+1}{2}\right)^2+b=\left(b^2-a^2+1\right)^2+b\)

\(\Leftrightarrow-\left(a-b\right)\left(a^3+a^2b-ab^2-2a-b^3-2b-1\right)=0\Rightarrow a=b\)

\(\Leftrightarrow\sqrt{x+y+1}=\sqrt{3x+3y}\Leftrightarrow x+y+1=3x+3y\Leftrightarrow2\left(x+y\right)=1\Leftrightarrow y=\dfrac{1-2x}{2}\)

\(với:y=\dfrac{1-2x}{2}\Rightarrow pt\) \(dưới\)

\(\Leftrightarrow6x\left(-10x^2+x+3\right)=-1-3\left(2x-1\right)^2\left(3+5x\right)\)

\(rút\) \(gọn\Leftrightarrow-\left(3x-2\right)\left(6x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\Rightarrow y=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\Rightarrow y=\dfrac{4}{3}\end{matrix}\right.\)(thỏa)

bài này chắc có cánh làm đạo hàm f(t') để ra cái a=b nhưng mình chưa học nên phân tích thủ công tí