\(3x^2+y^2+10x-2xy+26=0\)

\(\left(x^2-2xy+y^2\right)+2.\left(x^2+2.2,5x+2,5^2\right)+19,75=0\)

\(\left(x-y\right)^2+2.\left(x+2,5\right)^2+19,75=0\)(1)

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x;y\\2.\left(x+2,5\right)^2\ge0\forall x\end{cases}\Rightarrow\left(x-y\right)^2+2.\left(x+2,5\right)^2+19,75\ge19,75}\)

\(\Rightarrow\left(x-y\right)^2+2.\left(x+2,5\right)^2+19,75>0\forall x;y\)(2)

Từ (1) và (2)

\(\Rightarrow\)x;y không có giá trị

Vậy x;y không có giá trị

a)  \(49x^2-56x+16\) 

\(=\left(7x-4\right)^2\)

\(=\left(7.2-4\right)^2=100\)

b) mk chỉnh lại đề

  \(27x^3+54x^2+36x+8\)  

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

\(=\left[3.\left(-2\right)+2\right]^3=-64\)

c)  \(\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x+1\right)\left(x^2+x+1\right)+3\left(x-1\right)^2\)

\(=6x^2+7x+5\)

\(=6.\left(-\frac{2}{5}\right)^2+7.\left(-\frac{2}{5}\right)+5\)

\(=\frac{79}{25}\)

Bạn ơi có đúng không?

\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)

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

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

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

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

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

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

\(=x^2\left(x-1\right)\left(x-1\right)^2\)

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

\(a,=\left(2x-7\right)^2=\left(2.4-7\right)^2=1\)

\(b,\left(x-3\right)^3=\left(5-3\right)^3=8\)

a)   \(4x^2-28x+49\)

\(=\left(2x-7\right)^2\)

\(=\left(2.4-7\right)^2=1\)

b)  \(x^3-9x^2+27x-27\)

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

\(=\left(5-3\right)^3=8\)

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

b)  \(27y^3-9y^2+y-\frac{1}{27}=\left(3y-\frac{1}{3}\right)^3\)

c) \(8x^6+12x^4y+6x^2y+y^3=\left(2x^2+y\right)^3\)

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

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