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

\(\Delta>0\Leftrightarrow1-\left(-4m\right)>0\Leftrightarrow m>-\dfrac{1}{4}\)(thì P (d) cắt tại 2 đ pb)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=1\\x1.x2=-m\end{matrix}\right.\) \(\Rightarrow A\left(x1;y1\right),B\left(x2;y2\right)\)

\(I\) \(là\) \(tđ\) \(AB\Rightarrow I\left(\dfrac{x1+x2}{2};\dfrac{y1+y2}{2}\right)=\left(\dfrac{1}{2};\dfrac{x1+x2+2m}{2}\right)=\left(\dfrac{1}{2};\dfrac{2m+1}{2}\right)\)

\(\Rightarrow OI=\sqrt{\dfrac{13}{2}}=\sqrt{\left(0-\dfrac{1}{2}\right)^2+\left(0-\dfrac{2m+1}{2}\right)^2}\)

\(\Leftrightarrow\dfrac{13}{2}=\dfrac{1}{4}+\dfrac{\left(2m+1\right)^2}{4}\Leftrightarrow\left[{}\begin{matrix}m=2\left(tm\right)\\m=-3\left(ktm\right)\end{matrix}\right.\)

\(x^2=mx-m+5\Leftrightarrow x^2-mx+m-5=0\left(1\right)\)

\(\Delta>0\Leftrightarrow m^2-4\left(m-5\right)>0\Leftrightarrow m^2-4m+20>0\left(đúng\forall m\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=m\\x1x2=m-5\end{matrix}\right.\)

\(\dfrac{1}{x1}+\dfrac{1}{x2}=-\dfrac{3}{2}\left(x1;x2\ne0\right)\Rightarrow\dfrac{x1+x2}{x1x2}=-\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{m}{m-5}=-\dfrac{3}{2}\Leftrightarrow m=3\) \(m=3\Rightarrow x_{12}=\dfrac{3\pm\sqrt{17}}{2}\left(tm\right)\Rightarrow m=3\left(tm\right)\)

đoạn này bị sai chút \(\Sigma\dfrac{a}{ab+1}\ge3-\dfrac{1}{2}\left(a\sqrt{ab}+b\sqrt{bc}+c\sqrt{ca}\right)=3-\dfrac{1}{2}\Sigma\sqrt{a^3b}\)

có: \(BĐT:\left(x+y+z\right)^2\ge3\left(x^3y+y^3z+z^3z\right)\)

\(\Rightarrow\Sigma\dfrac{a}{ab+1}\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)

đề thế này à bạn nãy mình đọc chắc nhầm

\(\Sigma\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge\Sigma a-\dfrac{a^2b}{2\sqrt{ab}}=\Sigma a-\dfrac{a\sqrt{ab}}{2}\)

\(\Rightarrow\Sigma\dfrac{a}{ab+1}\ge a+b+c-(\dfrac{a\sqrt{ab}}{2}+\dfrac{b\sqrt{bc}}{2}+\dfrac{c\sqrt{ca}}{2})\)

\(=3-\dfrac{1}{2}\left(\Sigma a\sqrt{ab}\right)\ge3-\dfrac{1}{2}\left(\dfrac{a\left(a+b\right)}{2}+\dfrac{b\left(b+c\right)}{2}+\dfrac{c\left(c+a\right)}{2}\right)=3-\dfrac{1}{4}\left(a^2+b^2+c^2+ab+bc+ca\right)=3-\dfrac{1}{4}\left[\left(a+b+c\right)^2-\left(ab+bc+ca\right)\right]\ge3-\dfrac{1}{4}\left[3^2-\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{1}{4}\left[3^2-\dfrac{3^2}{3}\right]=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow a=b=c=1\)

\(B=\Sigma\dfrac{1}{b+1}\ge\dfrac{9}{a+b+c+1+1+1}=\dfrac{9}{3+3}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow a=b=c=1\)

\(A=\Sigma\dfrac{x^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{x+y+y+z+x+z}=\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{3^2}{2.3}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow x=y=z=1\)

\(y=\sqrt{x^2-\left(2m+1\right)x+m-2}\)

\(f\left(x\right)=x^2-\left(2m+1\right)x+m-2\ge0\) đúng với \(x\in\left[-1;2\right]\)

\(TH1:\Delta\le0\Leftrightarrow\left(2m+1\right)^2-4\left(m-2\right)\le0\Leftrightarrow4m^2+9\le0\)(loại)

\(TH2:\)\(\Delta'>0\Leftrightarrow\)\(4m^2+9>0\)(luôn đúng)

\(\Leftrightarrow\left[{}\begin{matrix}2\le x1< x2\left(1\right)\\x1< x2\le-1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}1f\left(2\right)\ge0\\\dfrac{S}{2}>2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2^2-\left(2m+1\right).2+m-2\ge0\\2m+1>4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\le0\\m>\dfrac{3}{2}\end{matrix}\right.\)(loại)

\(\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}1f\left(-1\right)\ge0\\\dfrac{S}{2}< -1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-\left(2m+1\right)\left(-1\right)+m-2\ge0\\2m+1< -2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\m< -\dfrac{3}{2}\end{matrix}\right.\)(vô lí) vậy ko có giá trị nào của m thỏa

đề này lạ nhỉ bth là tam thức bậc 2 thường có -2(m+5)x nữa

chứ chả ai cho đề kiểu này làm b=0 đâu nhỉ

\(f\left(x\right)=3x^2-2\left(m+5\right)x-m^2+2m+8\le0\) (1) nghiệm đúng với mọi x thuộc [-1;1]

\(\Leftrightarrow x1\le-1< 1\le x2\Leftrightarrow\left\{{}\begin{matrix}3f\left(-1\right)\le0\\3f\left(1\right)\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\left(m+5\right)-m^2+2m+11\le0\\-2\left(m+5\right)-m^2+2m+11\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\le-3\\m\ge7\end{matrix}\right.\\\left[{}\begin{matrix}m\le-1\\m\ge1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m\le-3\\m\ge7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+4x+y^3+3=0\\x^2y^3+y=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2+2\left(x^2y^3+y\right)+y^3+3=0\left(1\right)\\x^2y^3+y=2x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x^2+2x^2y^3+2y+y^3+3=0\Leftrightarrow\left(y+1\right)\left(2x^2y^2-2x^2y+y^2+2x-y+3\right)=0\)

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

\(\left(3\right)\Leftrightarrow2x^2\left(y^2-y+2\right)+y^2-y+3=0\)

\(\Rightarrow a=y^2-y+2=\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)

\(\Delta=0-4.2\left(y^2-y+2\right)\left(y^2-y+3\right)=-8\left(y^2-y+2\right)\left(y^2-y+3\right)\)

\(y^2-y+3=\left(y-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\)

\(\Rightarrow\Delta=-8\left(y^2-y+2\right)\left(y^2-y+3\right)< 0\)

\(\Rightarrow\left(3\right)\) không tồn tại nghiệm (x;y) nào

do đó hpt có  nghiệm x=y=-1

\(P\ge\dfrac{\left(x+y+z\right)^2}{\sqrt{15x^2+26xy+8y^2}+\sqrt{15y^2+26yz+8z^2}+\sqrt{15z^2+26zx+8x^2}}\)

\(\sqrt{15x^2+26xy+8y^2}=\sqrt{\left(5x+2y\right)\left(3x+4y\right)}\le\dfrac{\left(5x+2y+3x+4y\right)}{2}=\dfrac{8x+6y}{2}\)

\(tt\Rightarrow\sqrt{15y^2+26yz+8z^2}\le\dfrac{8y+6z}{2};\sqrt{15z^2+26zx+8x^2}\le\dfrac{8z+6x}{2}\)

\(\Rightarrow P\ge\dfrac{3^2}{\dfrac{8x+6y+8y+6z+8z+6x}{2}}=\dfrac{18}{14\left(x+y+z\right)}=\dfrac{18}{14.3}=\dfrac{3}{7}\)

\(dâu"="\Leftrightarrow x=y=z=1\)