\(A=\left(4x^2+25y^2+9-20xy-12x+30y\right)+\left(9x^2+6x+1\right)-2\)

\(A=\left(2x-5y-3\right)^2+\left(3x+1\right)^2-2\ge-2\)

\(A_{min}=-2\) khi \(\left\{{}\begin{matrix}x=-\frac{1}{3}\\y=-\frac{11}{15}\end{matrix}\right.\)

\(B=\left(x^2-3x+\frac{9}{4}\right)+\left(y^2-4y+4\right)-\frac{8105}{4}\)

\(B=\left(x-\frac{3}{2}\right)^2+\left(y-2\right)^2-\frac{8105}{4}\ge-\frac{8105}{4}\)

\(B_{min}=-\frac{8105}{4}\) khi \(\left\{{}\begin{matrix}x=\frac{3}{2}\\y=2\end{matrix}\right.\)

\(a,=\left(x-2\right)\left(9x^2y^2-6x^3y^2\right)=3x^2y^2\left(3-2x\right)\left(x-2\right)\\ b,=5x\left(x^2-y^2\right)+20x\left(x+y\right)=5x\left(x-y\right)\left(x+y\right)+20x\left(x+y\right)\\ =5\left(x+y\right)\left(x^2-xy+4x\right)\\ c,=8x^2+2x-12x-3=2x\left(4x+1\right)-3\left(4x+1\right)=\left(2x-3\right)\left(4x+1\right)\)

B= \(\frac{3y^2}{-25x^2+20xy-5y^2}=\frac{3y^2}{-y^2-\left(25x^2-20xy+4y^2\right)}=\frac{1}{-\frac{y^2}{3y^2}-\frac{\left(5x-2y\right)^2}{3y^2}}\)

=\(\frac{1}{-\frac{1}{3}-\frac{\left(5x-2y\right)^2}{3y^2}}\)

Có \(\frac{1}{3}+\frac{\left(5x-2y\right)^2}{3y^2}\ge\frac{1}{3}\) vs mọi x,y và y\(\ne0\)

<=>\(-\frac{1}{3}-\frac{\left(5x-2y\right)^2}{3y^2}\le-\frac{1}{3}\)

<=> \(\frac{1}{-\frac{1}{3}-\frac{\left(5x-2y\right)^2}{3y^2}}\ge-3\) <=> B \(\ge3\)

Dấu "=" xảy ra <=> 5x-2y=0

<=> 5x=2y < => \(x=\frac{2y}{5}\)

Vậy minB=3 <=> \(x=\frac{2y}{5}\)

Ta có: \(P=2x^3+10x^2-6x+7;Q=-2x^3-10x^2+6x-7+2x^2=-P+2x^2\)

Như vậy \(P+Q=2x^2\ge0.\)

Nếu P và Q cùng âm thì ta thấy ngay \(P+Q< 0\)(Vô lý)

Vậy P và Q không thể cùng âm.

Chúc em luôn học tập tốt :)))

\(A=\frac{6}{2x^2-6x+7}\)

Ta có \(2x^2-6x+7=2\left(x^2-3x+\frac{9}{4}\right)+\frac{5}{2}=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

\(\Rightarrow A=\frac{6}{2x^2-6x+7}\le\frac{6}{\frac{5}{2}}=\frac{12}{5}\)

\(Max_A=\frac{12}{5}\Leftrightarrow x=\frac{3}{2}\)