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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2-4xy+2x-4y+3+4y^2\)
\(A=x^2-2.2xy+\left(2y\right)^2+2x-4y+3\)
\(A=\left(x-2y\right)^2-2.\left(x-2y\right)+1+2\)
\(A=\left(x-2y-1\right)^2+2\ge2\)
Vậy GTNN của A=2.
![](https://rs.olm.vn/images/avt/0.png?1311)
c/\(x^2-2x=2y-xy\)
\(x^2-2x+xy-2y=0\)
\(x\left(x-2\right)+y\left(x-2\right)=0\)
\(\left(x+y\right)\left(x-2\right)=0\)
d/\(x^2+4xy-16+4y^2\)
\(=\left(x^2+4xy+4y^2\right)-16\)
\(=\left(x+2y\right)^2-4^2\)
\(=\left(x+2y-4\right)\left(x+2y+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\frac{16}{x-3}+2009\)
\(=x-3+\frac{16}{x-3}+2009\)
\(\ge2\sqrt{\left(x-3\right)\cdot\frac{16}{x-3}}+2009\)
\(=8+2009=2017\)
Dấu "=" xảy ra tại x=7
\(B=\left(x-3\right)+\frac{16}{x-3}++2012\)
\(\ge2\sqrt{\frac{16\left(x-3\right)}{x-3}}+2012\)
\(=8+2012=2020\)
(Dấu "="\(\Leftrightarrow x=7\))