a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

bạn viết lại đi sai đề hay sao ý

a: A=2/3x^2y+4x^2y=14/3x^2y

=14/3*9*7=294

b: B=xy^2(1/2+1/3+1/6)=xy^2=3/4*1/4=3/16

c: C=x^3y^3(2+10-20)=-8x^3y^3

=-8*1^3(-1)^3=8

d: D=xy^2(2018+16-2016)

=18xy^2

=18(-2)*1/9=-4

m: (x-y)(x^2-2xy+y^2)

=(x-y)*(x-y)^2

=(x-y)^3

=x^3-3x^2y+3xy^2-y^3

n: =-(x^3+x^2y-x-x^2y-xy^2+y)

=-x^3+x+xy^2-y

o: =-(x^3+x^2y^2-x^2-2xy-2y^3+2y)

=-x^3-x^2y^2+x^2+2xy+2y^3-2y

p: (1/2x-1)(2x-3)

=1/2x*2x-1/2x*3-2x+3

=x^2-3/2x-2x+3

=x^2-7/2x+3

q: (x-1/2y)(x-1/2y)

=(x-1/2y)^2

=x^2-xy+1/4y^2

r: (x^2-2x+3)(1/2x-5)

=1/2x^3-5x^2-x^2+10x+3/2x-15

=1/2x^3-6x^2+11,5x-15