a)

\(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

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

Vì \(\hept{\begin{cases}\left(2x-3y\right)^2\ge0\\\left(x-3\right)^2\ge0\end{cases}}\)nên

\(\Rightarrow\hept{\begin{cases}\left(2x-3y\right)^2=0\\\left(x-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

Vậy x=3 và y=2

b)

\(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

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

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(x-5\right)^2\ge0\\\left(y-4\right)^2\ge0\end{cases}}\)nên

\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(x-5\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\x-5=0\\y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=0\\x=5\\y=4\end{cases}}}\)( VÔ nghiệm vì \(x+y\ne0\))

Vậy không có giá trị x, y nào thỏa mãn đề bài

Bài này giải rồi mà

a , \(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow25x^2+45y^2-60xy-30x+45=0\)

\(\Leftrightarrow\left(5x\right)^2-2.5.\left(6y+3\right)+\left(6y+3\right)^2+9y^2-36y+36=0\)

\(\Leftrightarrow\left(5x-6y-3\right)^2+9\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(5x-6y-3\right)^2+9\left(y-2\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(5x-6y-3\right)^2\ge0\\9\left(y-2\right)^2\ge0\end{matrix}\right.\Rightarrow\left(5x-6y-3\right)^2+9\left(y-2\right)^2\ge0\)

Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}5x-6y-3=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Vậy ...

a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng

Bài 1 : 

\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)

\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)

\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)

\(\Leftrightarrow\)\(9x=16\)

\(\Leftrightarrow\)\(x=\frac{16}{9}\)

Vậy \(x=\frac{16}{9}\)

Chúc bạn học tốt ~ 

      \(5x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow\left(4x^2+9y^2-12xy\right)+\left(x^2-6x+9\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3y\\x=3\end{cases}\Rightarrow}\hept{\begin{cases}y=2\\x=3\end{cases}}}\)

\(5x^2+9y^2-12xy-6x+9=0\)

<=>  \(\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

<=>  \(\left(2x-3y\right)^2+\left(x-3\right)^2=0\)

<=>  \(\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}}\)

<=>  \(\hept{\begin{cases}y=2\\x=3\end{cases}}\)

Vậy...