\(A=-x^2+2xy-4y^2+2x+10y-3\)

\(=-x^2+2xy-y^2+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10< =10\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y+1=3\end{matrix}\right.\)

\(B=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37< =37\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\)

=>x=y=5

\(M=18+4x-8y+6xy+5x^2+10y^2\)

\(=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)

\(=\left(x+y\right)^2+4\left(x+\frac{1}{2}\right)^2+\left(y-4\right)^2+1\)

Có \(\left(x+y\right)^2\ge0\forall xy\)

\(4\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\left(y-4\right)^2\ge0\forall y\)

\(\Rightarrow M\ge1\forall x,y\)

hay \(M>0\forall x,y\)

\(M=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)

\(M=\left(x+3y\right)^2+\left(2x+1\right)^2+\left(y-4\right)^2+1>0;\forall x;y\)

Ta có: \(M=18+4x-8y+6xy+5x^2+10y^2\)

\(=4x^2+4x+1+x^2+6xy+9y^2+y^2-8y+16+1\)

\(=\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\)

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\left(x+3y\right)^2\ge0\forall x,y\)

\(\left(y-4\right)^2\ge0\forall y\)

Do đó: \(\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\ge1>0\forall x,y\)

hay \(M>0\forall x,y\)

Ta có : 3x² - 7xy + 4y² = 0

=> 3x² - 3xy - 4xy + 4y² = 0

=> 3x( x - y) - 4y( x - y) = 0

=> ( x - y)( 3x - 4y) = 0

=> x = y ; 3x = 4y

Thay : x = y ; 3x = 4y vào phân thức trên ta có:

\(A=\dfrac{4y+2x}{5y-7x}+\dfrac{3x-2y}{10y-4x}\)

\(A=\dfrac{3x+2x}{5x-7x}+\dfrac{4y-2y}{10x-4x}\)

\(A=\dfrac{5x}{-2x}+\dfrac{2y}{6x}=\dfrac{5}{-2}+\dfrac{1}{3}=\dfrac{-13}{6}\)

\(10x+10y=10\left(x+y\right)=10.\left(-2\right)=-20\)

Q=10(x+y)=-2x10=-20

Đáp án : -80

Nếu cần cách giải nx thì ns tui nha

Học tốt

\(x^2-4x+5y^2-10y+9=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(5y^2-10y+5\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y^2-2y+1\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y-1\right)^2=0\)

Vì \(\left(x-2\right)^2\ge0;5\left(y-1\right)^2\ge0\) mà \(\left(x-2\right)^2+5\left(y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\5\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)