Đề bài là phân tích thành nhân tử đúng ko? chứ đưa về hđt thì đây là dạng của một hằng đẳng thức rồi còn gì.Ta có hằng đẳng thức \(a^2-b^2=\left(a-b\right)\left(a+b\right)\).Áp dụng vào:

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

\(=\left(-2x-4\right)\left(6x+2\right)\)

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

\(x^2+10x+26+y^2+2y\)

\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

\(\left(x+y+4\right)\left(x+y-4\right)\)

\(=\left(x+y\right)^2-16\)

\(=x^2+y^2+2xy-16\)

a, =(x^2 +10x+25) +(y^2 +2y+1)

    = (x+5)^2 +(y+1)^2

b, =(x+y)^2 -4^2

    = x^2 + 2xy+ y^2 -16

\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\) 

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

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

\(B=9x^2-\left(y^2-4y+4\right)\) 

   \(=9x^2-\left(y-2\right)^2\) 

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

A = ( 6x - 3y ) + (4x² - 4xy + y² )

A = 3.( 2x - y) + [ ( 2x )² - 2.2.x.y + y² ]

A = 3.( 2x - y ) + ( 2x - y )²

A = ( 2x - y ).(3 + 2x - y )

B = 9x² - ( y² - 4y + 4 )

B = ( 3x )² - ( y - 2 )²

B = ( 3x - y + 2 ).( 3x + y - 2 )

C = - 25x² + y² - 6y + 9

C =   ( y² - 2.3.y + 3² ) - ( 5x )²

C = ( y - 3 )² - ( 5x )²

C = (y - 3 - 5x ).( y - 3 +5x )

D = x² - 4x - y² -- 8y  - 12

D = ( x² - 4x + 4 ) - 4 - y² - 8y -12

D = ( x - 2.2x + 2² ) - ( y² + 2.4.y + 4² )

D = ( x - 2 )² - ( y + 4 )²

D = ( x - 2 + y + 4 ).( x - 2 - y - 4 )

D = ( x + y + 2 ).( x - y - 6 )

a, \(\left(4x+5\right)^2=\left(4x+5\right)\left(4x+5\right)=\left[\left(4x+5\right)4x\right]+\left[\left(4x+5\right)5\right]=4x^2+20x+25\)

b, \(\left(5x-2\right)^2=\left(5x-2\right)\left(5x-2\right)=\left[\left(5x-2\right)5x-\left(5x-2\right)2\right]=5x^2-10x+25\)

b, \(8^2-12x^2=\left(8^2-12x^2\right)\left(8^2+12x^2\right)\)

đúng ko :) 

@No name: Bị sai rồi nhé, a,b,c sai hết  :>

a) ( 4x + 5 )² 

= ( 4x )² + 2.4x.5 + 5² 

= 16x² + 40x + 25 

b) ( 5x - 2 )² 

= ( 5x )² - 2.5x.2 + 2² 

= 25x² - 20x + 4 

c) 8² - 12x² 

= 64 - 12x² 

= ( V8 - V12x )( V8 + V12x ) 

a)\(-25+4x^2=\left(2x-5\right)\left(2x+5\right)\)

b)\(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

c)\(\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)

\(a,-25+4x^2=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(b,-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

\(c,\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x\right)^2+\frac{2.1}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)(sửa đề)

Bài giải

\(a,\text{ }a^2+9-6a=a^2+2\cdot3a+3^2=\left(a-3\right)^2\)

\(b,\text{ }x^2-x+\frac{1}{4}=x^2-2\cdot\frac{1}{2}\cdot x+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)

\(c,\text{ }-x^2+4x-x=3x-x^2=\left(\sqrt{3x}\right)^2-x^2=\left(\sqrt{3x}-x\right)\left(\sqrt{3x}+x\right)\)( Đề nói vận dụng hằng đẳng thức để rút gọn nên mình đưa về hiệu hai ình phương nha ! )

a) \(a^2+9-6a\)

\(=\left(a-3\right)^2\)

b) \(x^2-x+\frac{1}{4}\)

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

c) \(-x^2+4x-x\)

\(=3x-x^2\)

\(=x\left(3-x\right)\)

Ta có : \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)

    \(\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

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

Do \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\\\left(y-3\right)^2\ge0\\\left(z-5\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=y+z\\y=3\\z=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}}\)

Khi đó \(P=\left(4-4\right)^{2018}+\left(3-4\right)^{2018}+\left(5-4\right)^{2018}\)

               \(=0+\left(-1\right)^{2018}+1^{2018}\)

               \(=2\)