1:

\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)

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

\(\Delta'=\left(m+1\right)^2-\left(2m-3\right)=m^2+4>0\) ; \(\forall m\)

\(\Rightarrow\) Phương trình luôn có 2 nghiệm pb với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=2m-3\end{matrix}\right.\)

Ta có: \(P=\left|\dfrac{x_1+x_2}{x_1-x_2}\right|\ge0\)

\(\Rightarrow P_{min}=0\) khi \(x_1+x_2=0\Leftrightarrow m=-1\)

Đề là yêu cầu tìm max hay min nhỉ? Min thế này thì có vẻ là quá dễ

\(A=\dfrac{2\left(x^3+y^3\right)}{\left(x^4+y^2\right)\left(x^2+y^4\right)}=2.\dfrac{\left(x^3+y^3\right)}{x^4y^4+x^2y^2+x^6+y^6}\)

\(=2.\dfrac{\left(x^3+y^3\right)}{1+1+x^6+y^6}=2.\dfrac{x^3+y^3}{x^6+y^6+2x^3y^3}=2.\dfrac{x^3+y^3}{\left(x^3+y^3\right)^2}=\dfrac{2}{x^3+y^3}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^3+y^3+1\ge3\sqrt{xy.1}=3\)

\(\Rightarrow x^3+y^3\ge2\Rightarrow\dfrac{2}{x^3+y^3}\le1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A\le1\)

Dấu "=" xảy ra khi x=y=1.

Vậy MaxA là 1, đạt được khi x=y=1.

\(\left(x^2;y^2\right)=\left(a;b\right)\Rightarrow P=\dfrac{\left(a-b\right)\left(1-ab\right)}{\left(1+a\right)^2\left(1+b\right)^2}\)

Ta có:

\(\left(a+b\right)\left(1+ab\right)-\left(a-b\right)\left(1-ab\right)=2b\left(a^2+1\right)\ge0;\forall a;b\ge0\)

\(\Rightarrow\left(a+b\right)\left(1+ab\right)\ge\left(a-b\right)\left(1-ab\right)\)

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

\(P_{max}=\dfrac{1}{4}\) khi \(\left(a;b\right)=\left(1;0\right)\) hay \(\left(x;y\right)=\left(1;0\right)\)

\(P=\dfrac{\left[\left(x-y\right)\left(1+xy\right)\right]\left[\left(x+y\right)\left(1-xy\right)\right]}{\left(1+x^2\right)^2\left(1+y^2\right)^2}\)

Áp dụng BĐT Cosi ta có:

\(\left(x-y\right)\left(1+xy\right)\le\dfrac{\left(x-y\right)^2+\left(1+xy\right)^2}{2}=\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{2}\\ \left(x+y\right)\left(1-xy\right)\le\dfrac{\left(x+y\right)^2+\left(1-xy\right)^2}{2}=\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{2}\)

\(\to P\le\dfrac{\left(1+x^2\right)^2\left(1+y^2\right)^2}{4\left(1+x^2\right)^2\left(1+y^2\right)^2}=\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\ge\dfrac{5}{4}\)

nên \(\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\right]^2\ge\dfrac{25}{16}\)

Dấu '=' xảy ra khi x=-1/2

Có \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\ge\dfrac{5}{4}\forall x\)

\(A=\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\right]^2\ge\left(\dfrac{5}{4}\right)^2=\dfrac{25}{16}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy min \(A=\dfrac{25}{16}\Leftrightarrow x=\dfrac{-1}{2}\)

\(1.x^2+\dfrac{1}{x^2}-2m\left(x+\dfrac{1}{x}\right)+1+2m=0\left(1\right)\)\(đặt:x^2+\dfrac{1}{x^2}=t\)

\(x>0\Rightarrow t\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)

\(x< 0\Rightarrow-t=-x^2+\dfrac{1}{\left(-x^2\right)}\ge2\Rightarrow t\le-2\)

\(\Rightarrow t\in(-\infty;-2]\cup[2;+\infty)\left(2\right)\)

\(\Rightarrow\left(1\right)\Leftrightarrow t^2-2mt+2m-1=0\)

\(\Leftrightarrow\left(t-1\right)\left(t-2m+1\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\notin\left(2\right)\\t=2m-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2m-1\le-2\\2m-1\ge2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\le-\dfrac{1}{2}\\m\ge\dfrac{3}{4}\end{matrix}\right.\)

\(2.\)  \(f^2\left(\left|x\right|\right)+\left(m-2\right)f\left(\left|x\right|\right)+m-3=0\left(1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}f\left(\left|x\right|\right)=-1\\f\left(\left|x\right|\right)=3-m\end{matrix}\right.\)

\(dựa\) \(vào\) \(đồ\) \(thị\) \(f\left(\left|x\right|\right)\) \(\Rightarrow f\left(\left|x\right|\right)=-1\) \(có\) \(2nghiem\) \(pb\)

\(\left(1\right)có\) \(6\) \(ngo\) \(pb\Leftrightarrow\left\{{}\begin{matrix}-1< 3-m< 3\\3-m\ne-1\\\end{matrix}\right.\)\(\Leftrightarrow0< m< 4\)

\(\Rightarrow m=\left\{1;2;3\right\}\)

\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)

\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)

\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)

\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)

\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)

Vậy \(P=\dfrac{37}{60}\)

\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)

\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)

\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)

\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)

Vậy \(Q=\dfrac{29}{125}\)