là làm sao ạ ???

\(x^2-6x+y^2-2y+12\)

= \(\left(x^2-6x+9\right)+\left(y^2-2y+1\right)+2\)

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

mà \(\left(x-3\right)^2\) ≥ 0 ∀ x và \(\left(y-1\right)^2\)≥ 0 ∀ y

=> Min A = 2 <=> \(\left(x-3\right)^2\)=0 <=> x=3

và (y-1)\(^2\)=0 <=> y=1

vậy Min A =2 khi x=3 và y=1

b, \(x^2+3x-y^2+3y\)

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

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

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

c,\(3\left(x+3\right)-x^2+9\)

=\(3\left(x+3\right)-\left(x^2-9\right)\)

=3(x+3)-(x+3)(x-3)

=(x+3)(3-x+3)

=(x+3)x

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

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

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

A=y(x-1)(x+1)+x(y-1)(y+1)

thay x=5 và y=2 ta có:

A=2(5-1)(5+1)+5(2-1)(2+1)

A=2.4.6+5.1.3

A=63

mk nhìn đề bài mà choáng á!!!!

1,\(6x^2-9\)

=\(\left(4x^{ }\right)^2-3^2\)

=(4x-3)(4x+3)

2,\(x^2-4+\left(x+2\right)^2\)

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

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

=(x+2)(x-2+x+2)

=2x(x+2)

3,7(x-5)+8x(5-x)

=7(x-5)-8x(x-5)

=(x-5)(7-8x)

4,\(2x^2-18\)

=2\(\left(x^2-9\right)\)

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

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

5,2x\(^2\)-50

=2(x\(^2\)-25)

=2(\(x^2-5^2\))

=2(x+5)(x-5)

sai chỗ nào thì bạn bảo mk nhé!

6x(x-1)=x-1

6x=(x-1):(x-1)

6x=1

x=\(\frac{1}{6}\)

vậy x=\(\frac{1}{6}\)