`3/5 : 1/6 + 3:5 + 6/10 xx12 + 12/20`

`=3/5 xx 6 + 3/5 + 3/5 xx12 + 3/5`

`=3/5 xx (6+1+12+1)`

`=3/5 xx20`

`=12`

`5/8 - (-2/5) - 3/10`

`=5/8  +2/5 - 3/10`

`=25/40 +16/40 - 12/40`

`=(25+16-12)/40`

`=29/40`

Thay `x=2, y=-1` vào `A` ta có:

`A=x^2 y -xy^2 +xy`

`=2^2 . (-1) - 2 . (-1)^2 + 2.(-1)`

`=4.(-1)-2.1+(-2)`

`=-4-2-2`

`=-8`

Thay `x=2, y=-1` vào `B` ta có:

`B=xy(x-y+1)`

`=2.(-1)[2-(-1)+1]`

`=-2.4`

`=-8`

Ta có:
`2 sqrt(3) = sqrt(4) . sqrt(3) = sqrt(4.3) = sqrt(12)`

`3 sqrt(2) = sqrt(9) . sqrt(2) = sqrt(9.2) = sqrt(18)`

Vi `sqrt(12) <sqrt(18) => 2 sqrt(3) < 3sqrt(2)`

\(5\sqrt{\dfrac{2}{9}}+3\sqrt{\dfrac{2}{25}}-15\sqrt{\dfrac{2}{81}}\\ =\dfrac{5\sqrt{2}}{\sqrt{9}}+\dfrac{3\sqrt{2}}{\sqrt{25}}-\dfrac{15\sqrt{2}}{\sqrt{81}}\\ =\dfrac{5\sqrt{2}}{3}+\dfrac{3\sqrt{2}}{5}-\dfrac{15\sqrt{2}}{9}\\ =\dfrac{5\sqrt{2}}{3}+\dfrac{3\sqrt{2}}{5}-\dfrac{5\sqrt{2}}{3}\\ =\dfrac{3\sqrt{2}}{5}\)

Ta có:

`x^2 >= 0`

`-4/25 < 0`

Mà đề cho `x^2 = -4/25`(vô lí)

Vậy phương trình có tập nghiệm $S= \varnothing$

`c,x^2 +2y^2 -2xy+2x-6y+5=0`

`<=>x^2 +2x(1-y)+2y^2 -6y+5=0`

`<=>[x^2 +2x(1-y)+(1-y)^2]-(1-y)^2+2y^2 -6y+5=0`

`<=>(x+1-y)^2 -1+2y-y^2+2y^2 -6y+5=0`

`<=>(x+1-y)^2 +y^2 -4y+4=0`

`<=>(x+1-y)^2 +(y-2)^2=0`

Vì `(x+1-y)^2 > = 0` \(\forall x\) `(y-2)^2 >= 0` \(\forall y\)

`=> (x+1-y)^2 + (y-2)^2 >= 0` \(\forall x,y\)

Dấu "=" xảy ra `<=>{(x+1-y=0),(y-2=0):}<=>{(x+1-2=0),(y=2):}<=>{(x-1=0),(y=2):}<=>{(x=1),(y=2):}`

Vậy `(x,y)=(1,2)`

`b, -2x^2 +4x-y^2 -2y-3=0`

`<=>-2x^2 +4x-2 - y^2 -2y -1=0`

`<=>-2(x^2 -2x +1) - (y^2 +2y+1) = 0`

`<=>-2(x-1)^2 - (y+1)^2 = 0`

`<=>2(x-1)^2 + (y+1)^2 = 0`

Vì \(2\left(x-1\right)^2\ge0\forall x;\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow2\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra`<=>{(x-1=0),(y+1=0):}<=>{(x=1),(y=-1):}`

Vậy `(x,y)=(1,-1)`