Phân tích theo cách nhóm hạng tử nhé bạn ! 

a, xy + xz - 2y - 2z

= x(y + z) - 2(y + z)

= (x - 2)(y + x)

b, x² - 6xy + 9y² - 25z²

= (x - 3y)² - 25z²

= (x - 3y - 5z)(x - 3y + 5z) 

Ngoài ra bạn có thể hỏi mình để bổ sung kiến thức nâng cao !

= ( x - 3y )^2 - (5z)^2

= ( x - 3y - 5z ) ( x - 3y + 5z )

x² - 6xy - 25z² + 9y²

= (x - 3y)² - (5z)²

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

# Học tốt #

Câu 1: 3x² - 6xy + 3y² - 12z²

= 3(x² - 2xy + y² - 4z²)

= 3[(x² -2xy+y²)- (2z)²]

= 3[(x-y)² - (2z)²]

=3(x-y-2z)(x-y +2z)

Câu 2: x² - 6xy - 25z²+ 9y²

= x² - 2x.3y + (3y)² - (5z)²

= (x-3y)² - (5z)²

= (x-3y-5z)(x-3y+5z) 

\(4x^2+4x-9y^2+1\\ =\left(4x^2+4x+1\right)-9y^2\\ =\left(2x+1\right)^2-9y^2\\ =\left(2x+1\right)^2-\left(3y\right)^2\\ =\left[\left(2x+1\right)+3y\right]\left[\left(2x+1\right)-3y\right]\\ =\left(2x+1+3y\right)\left(2x+1-3y\right)\)

\(x^2-6xy+9y^2-25z^2\\ =\left(x^2-6xy+9y^2\right)-25z^2\\ =\left(x-3y\right)^2-25z^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left[\left(x-3y\right)+5z\right]\left[\left(x-3y\right)-5z\right]\\ =\left(x-3y+5z\right)\left(x-3y-5z\right)\)

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

cảm ơn

a)\(x^2-6xy+9y^2-25z^2=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(5z\right)^2\)

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

b)\(xyz+x^2yz-6yz=yz\left(x^2+x-6\right)=yz\left(x^2+3x-2x-6\right)\)

\(=yz\left[x\left(x+3\right)-2\left(x+3\right)\right]=yz\left(x-2\right)\left(x+3\right)\)

a.x²-6xy+9y²​-25z^​2

​= ( x^​2-6xy+9y²)-25z²

​= [x^​2-2x3y+(3y)²]-25z²​

​= (x-3y)²-25²

​= (x-3y+25)(x-3y-25)

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

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

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

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

Lời giải:

1. $xy+xz-2y-2z=(xy+xz)-(2y+2z)$

$=x(y+z)-2(y+z)=(x-2)(y+z)$

2. $x^2-6xy+9y^2)-25z^2$

$=(x-3y)^2-(5z)^2=(x-3y-5z)(x-3y+5z)$
3.

$3x^2-3y^2-12x+12y=(3x^2-3y^2)-(12x-12y)=3(x^2-y^2)-12(x-y)=3(x-y)(x+y)-12(x-y)=3(x-y)(x+y-4)$

Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{2}{x}+\frac{8}{9y}+\frac{18}{25z}\right)(x+y+z)\geq (\sqrt{2}+\sqrt{\frac{8}{9}}+\sqrt{\frac{18}{25}})^2\)

$\Leftrightarrow A.2\geq \frac{2312}{225}$

$\Leftrightarrow A\geq \frac{1156}{225}$

Vậy $A_{\min}=\frac{1156}{225}$