a. \(9x^2+25-12xy+5y^2-10y\)

\(=\left(9x^2-12xy+4y^2\right)+\left(25+y^2-10y\right)\)

\(=9\left(x^2-\frac{4xy}{3}+\frac{4y^2}{9}\right)+\left(5-y\right)^2\)

\(=9\left(x-\frac{2y}{3}\right)^2+\left(5-y\right)^2\)

a)( 6x - 2)²  ( 5x - 2)² - 2( 6x - 2 )( 5x - 2 ) 

=(6x-2)²-2(6x-2)(5x-2)+(5x-2)2
=[(6x-2)-(5x-2)]2
=(6x-2-5x+2)²

=X²
b) ( x² + 3x + 1)² - 2( x² + 3x + 1)( 3x + 1) + ( 9x² - 6x + 1)

=( x² + 3x + 1)2 - 2( x² + 3x + 1)( 3x + 1)+[(3x)²-2.3x.1+1²]

=( x² + 3x + 1)2 - 2( x² + 3x + 1)( 3x + 1)+(3x+1)²
=[( x² + 3x + 1)-(  3x + 1)]²
=( x² + 3x + 1- 3x - 1)2
=(x²)²
=x4
 

-3x⁴y + 6x³y - 3x²y

= -3x²y( x² - 2x + 1 )

= -3x²y( x - 1 )²

-3x⁴y+6x³y-3x²y

=-3x²y(x²-2xy+1)

=-3x²y(x-1)²

\(b\ne0\)

\(a-b=\frac{a}{b}\Rightarrow ab-b^2=a\Rightarrow a\left(b-1\right)=b^2=b^2-1+1=\left(b-1\right)\left(b+1\right)+1\)

\(\Rightarrow\left(b-1\right)\left(a-b-1\right)=1\)

=> (b-1)=(a-b-1)=1 => a=4; b=2 Hoặc

Cảm ơn bạn :33

Điều kiện là x lớn hơn 0 , x không bằng 4 và x không bằng 9 ạ Mong nhận đc câu trả lời ạ

\(a^2-2b+6b+b^2=-10\)

\(\Leftrightarrow a^2-2a+6b+b^2+10=0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\left(1\right)\)

Vì \(\hept{\begin{cases}\left(a-1\right)^2\ge0\forall a\\\left(b+3\right)^2\ge0\forall b\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+3\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=-3\end{cases}}}\)

\(L=\frac{x+y}{z}+1+\frac{y+z}{x}+1+\frac{x+z}{y}+1-3\)

\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)