1/

\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)

\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)

\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)

2/

\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)

\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)

\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)

a: \(H=6x^3y^4-2x^4y^2+3x^2y^2+5x^4y^2-A\cdot x^3y^4\)

\(=x^3y^4\left(6-A\right)+x^4y^2\left(5-2\right)+3x^2y^2\)

\(=\left(6-A\right)\cdot x^3y^4+x^4y^2\cdot3+3x^2y^2\)

Để H có bậc là 6 thì 6-A=0

=>A=6

b: Khi A=6 thì \(H=\left(6-6\right)\cdot x^3y^4+3x^4y^2+3x^2y^2\)

\(=3x^4y^2+3x^2y^2\)

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

\(x^2+1>1>0\forall x\ne0\)

\(x^2>0\forall x\ne0\)

\(y^2>0\forall y\ne0\)

Do đó: \(x^2y^2\left(x^2+1\right)>0\forall x,y\ne0\)

=>\(H=3x^2y^2\left(x^2+1\right)>0\forall x,y\ne0\)

=>H luôn dương khi x,y khác 0

\(TH1:\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{3}x^4y^3< 0\\-\dfrac{3}{5}x^3y^4< 0\\\dfrac{1}{2}xy^3>0\end{matrix}\right.\)

\(TH2:\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{3}x^4y^3>0\\-\dfrac{3}{5}x^3y^4>0\\\dfrac{1}{2}xy^3>0\end{matrix}\right.\)

\(TH3:\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{3}x^4y^3>0\\-\dfrac{3}{5}x^3y^4< 0\\\dfrac{1}{2}xy^3< 0\end{matrix}\right.\)

\(TH4:\left\{{}\begin{matrix}x< 0\\y>0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{3}x^4y^3< 0\\-\dfrac{3}{5}x^3y^4>0\\\dfrac{1}{2}xy^3< 0\end{matrix}\right.\)

Vậy ....

Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$

a: M=2(-2x-3xy^2+1)-3xy^2+1

=-4x-6xy^2+2-3xy^2+1

=-4x-9xy^2+3

b: Thay x=-2 và y=3 vào M, ta được:

M=2*(-2)-3*(-2)*3^2+1

=-4+1+6*9

=54-3

=51