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

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 ...

Câu b thì sao bạn?

a)

\(A=x^2-4x+1=x^2-2.2x+2^2-3\)

\(=(x-2)^2-3\)

Vì \((x-2)^2\geq 0, \forall x\Rightarrow A\geq 0-3=-3\)

Vậy GTNN của $A$ là $-3$ khi $x=2$

b) \(B=(x-2)(x-6)+7=x^2-6x-2x+12+7\)

\(=x^2-8x+19=(x^2-2.4x+4^2)+3\)

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

Vì \((x-4)^2\geq 0, \forall x\Rightarrow B\geq 0+3=3\)

Vậy GTNN của $B$ là $3$ khi $x=4$

c)

\(C=4x-x^2=4-(x^2-4x+4)=4-(x-2)^2\)

Vì \((x-2)^2\geq 0\Rightarrow C\leq 4-0=4\)

Vậy GTLN của $C$ là $4$ khi $x=2$

d) \(D=x^2-2x+y^2-4y+16=(x^2-2x+1)+(y^2-4y+4)+11\)

\(=(x-1)^2+(y-2)^2+11\)

Vì \((x-1)^2\geq 0; (y-2)^2\geq 0, \forall x,y\)

\(\Rightarrow D\geq 0+0+11=11\)

Vậy GTNN của $D$ là $11$ khi \(\left\{\begin{matrix} x-1=0\\ y-2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)

câu a sai đề         

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 ~ 

a)\(x^2+4y^2+6x-12y+18=0\)

\(\Leftrightarrow\left(x^2+2\cdot x\cdot3+9\right)+\left[\left(2y\right)^2-2\cdot2y\cdot3+9\right]=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\2y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-2\cdot x\cdot5+25\right)+\left(y^2-2.y.4+16\right)=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-5=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=5\\y=4\end{matrix}\right.\)(vô lý)

1. Ta có: 

\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)

Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)

2.

a) \(x^2+4y^2+6x-12y+18=0\)

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

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

Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)

Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)

b) \(2x^2+2y^2+2xy-10x-8y+41=0\)

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

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

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Rồi giải tương tự như trên