x=6 

thì phải

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow\left(x-2\right)^2+56>0\forall x\)

Vậy pt vô nghiệm 

(a-b)(b-a)^2

= (a-b) ( b^2 - 2ab + a^2 )

= ab^2 - 2ba^2 + a^3 - b^3 + 2ab^2 + ba^2

\(\left(a-b\right)\left(b-a\right)^2\)

\(=\left(a-b\right)\left(a-b\right)^2\)

\(=\left(a-b\right)^3\)

\(=a^3-3a^2b+3ab^2-b^3\)

= ( a + b + a - b )[ ( a + b )² - ( a + b )( a - b ) + ( a - b )² ] - 6a²b

= 2a( a² + 2ab + b² - a² + b² + a² - 2ab + b² ) - 6a²b

= 2a( a² + 3b² ) - 6a²b

= 2a( a² + 3b² - 3ab )

Sửa đề: \(x^2-2xy+y^2+3x-3y-10\)

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

\(=\left(x-y\right)\left(x-y+3\right)-10\)

Đặt \(x-y=t\)

\(\Rightarrow t.\left(t+3\right)-10\)

\(=t^2+3t-10\)

\(=t^2+5t-2t-10\)

\(=t\left(t+5\right)-2\left(t+5\right)\)

\(=\left(t+5\right)\left(t-2\right)\)

\(=\left(x-y+5\right)\left(x-y-2\right)\)

ta có :

\(A=-\left(x^2-x+\frac{1}{4}\right)-\left(2y^2-y+\frac{1}{8}\right)-\left(4z^2-z+\frac{1}{16}\right)+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(\frac{7}{16}-\left(x-\frac{1}{2}\right)^2-2\left(y-\frac{1}{4}\right)^2-4\left(z-\frac{1}{8}\right)^2\le\frac{7}{16}\)

Vậy GTLN của A là \(\frac{7}{16}\text{ khi }\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{4}\\z=\frac{1}{8}\end{cases}}\)

x^2y^2−2x^2y+2xy^2

ta có :

\(\hept{\begin{cases}\left(x+y\right)^2=x^2+2xy+y^2\text{ và }&2\left(x+y\right)\left(x-y\right)=2x^2-2y^2&\end{cases}}\)

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

\(4A=12x^2+12y^2+4z^2+20xy-12yz-12zx-8x-8y+12\)

\(=9x^2+9y^2+4z^2+18xy-12yz-12zx+2\left(x^2+y^2+4-4x-4y+2xy\right)+x^2+y^2-2xy+4\)

\(=\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\)

Dấu \(=\)khi \(\hept{\begin{cases}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\).

Vậy \(minA=1\)khi \(x=y=1,z=3\).

\(A=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}\left(x^2y^2+\frac{2}{3}xy-\frac{8}{3}x-\frac{8}{3}y\right)+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}y^2-\frac{16}{9}y-\frac{16}{9}]\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2-\frac{2y}{9}]+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2]+1\)

\(\Leftrightarrow A\ge1\Leftrightarrow MinA=1\)

Dấu '' = '' xảy ra khi:

\(\hept{\begin{cases}z-\frac{3}{2}x-\frac{3}{2}y=0\\y-1=0\\x+\frac{y}{3}-\frac{4}{3}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}z=0\\y=1\\x=1\end{cases}}\)