`a, x^2 -2x +y^2 +1=0`

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

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

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

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

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

Vậy `(x,y)=(1,0)`

Ta có:

`VP=(a+b)^3 - 3ab(a+b)

`=(a^3 + 3a^2 b + 3ab^2 + b^3) - (3a^2 b + 3ab^2)`

`=a^3 +3a^2 b +3ab^2 +b^3 -3a^2 b - 3ab^2`

` = a^3 + b^3=VT(đpcm)`

`3x^2 y +3x^2 y - 2x^2 y`

`=(3+3-2)x^2 y`

`=4x^2 y`

Thay `x=-2 , y=-1` vào biểu thức ta có:
`4x^2 y = 4.(-2)^2 . (-1) = -4.4=-16`

`a,` có bạn làm r

`b,` Ta có:
`Δ=b^2 -4ac`

`=[-(m+3)]^2 - 4.1.(2m+2)`

`=(m+3)^2 -4(2m+2)`

`=m^2 +6m+9-8m-8`

`=m^2 -2m+1`

`=(m-1)^2 >= 0`

Suy ra phương trình luôn có 2 nghiệm

Theo Vi-ét: `x_1 + x_2 = (-b)/a = (-[-(m+3)])/1 = m+3`

                   `x_1 x_2 = c/a = (2m+2)/1 = 2m+2`

\(x_1^2+x_2^2=13\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=13\\ \Leftrightarrow\left(m+3\right)^2-2\left(2m+2\right)-13=0\\ \Leftrightarrow m^2+6m+9-4m-4-13=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow m\left(m-2\right)+4\left(m-2\right)=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m-2=0\\m+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}m=2\\m=-4\end{matrix}\right.\)

`a,` Thay `m=2` vào phương trình ta có:

`x^2 -2(2-1)x-2.2-1=0`

`<=>x^2 -2.1x-4-1=0`

`<=>x^2 -2x-5=0`

Ta có:

`Δ=b^2 -4ac`

`=(-2)^2 -4.1.(-5)`

`=4+20`

`=24>0`

Suy ra phương trình luôn có 2 nghiệm phân biệt

\(\Rightarrow x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-2\right)+\sqrt{24}}{2.1}=\dfrac{2+2\sqrt{6}}{2}=1+\sqrt{6}\\ x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-2\right)-\sqrt{24}}{2.1}=\dfrac{2-2\sqrt{6}}{2}=1-\sqrt{6}\)

Vậy ...

`a, 1,24:0,5+3,76xx2`

`=1,24xx2+3,76xx2`

`=(1,24+3,76)xx2`

`=5xx2`

`=10`

`b,5,28:0,25+4,72xx4`

`=5,28xx4+4,72xx4`

`=(5,28+4,72)xx4`

`=10xx4`

`=40`