\(A=x^2+2xy+y^2-4x-4y+1\)

\(A=\left(x+y\right)^2-4\left(x+y\right)+1\)

\(A=3^2-4.3+1\)

\(A=-2\)

\(x^2+2xy+y^2-4x-4y+\)\(1\)

\(=\left(x^2+2xy+y^2\right)-\left(4x+4y\right)+1\)

\(=\left(x+y\right)^2-4\left(x+y\right)+1\)

Thay x+y = 1, ta có:

\(=3^2-4.3+1=-2\)

\(\text{•}A=x^2-6x+10=\left(x-3\right)^2+1\\ A=\left(103-3\right)^2+1=10001\\ \text{•}B=x^2+0,2x+1,01\\ B=\left(1,01\right)^2+0,2.1,01+1,01=2,2321\\ \text{•}C=x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\\ C=\left(3,5-3,25.2\right)\left(3,5+3,25.2\right)=-30\)

\(A=x^2-6x+10\)

\(=x^2-2.x.3+3^2+1\)

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

tại x=103, ta có:

\(\left(103-3\right)^2+1=10001\)

a) Theo đầu bài ta có:
\(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2=10\)
\(\Rightarrow\left(2-y\right)^2+y^2=10\)
\(\Rightarrow4+y^2-4y+y^2=10\)
\(\Rightarrow2y^2-4y=6\)
\(\Rightarrow2\left(y^2-2y\right)=6\)
\(\Rightarrow y\left(y-2\right)=3\)
Mà \(\hept{\begin{cases}y-\left(y-2\right)=2\\y+\left(y-2\right)=k\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}\\y-2=\frac{k-2}{2}\end{cases}}}\)( với k là hằng số )
\(\Rightarrow y\left(y-2\right)=\frac{k+2}{2}\cdot\frac{k-2}{2}\)
\(\Rightarrow\frac{\left(k+2\right)\left(k-2\right)}{4}=3\)
\(\Rightarrow k^2-4=12\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4;-4\)
- Nếu k = 4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=3\\x=2-y=-1\end{cases}\Rightarrow x^3+y^3=-1+27=26}\)
- Nếu k = -4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=-1\\x=2-y=3\end{cases}\Rightarrow x^3+y^3=27+-1=26}\)
Vậy x³ + y³ = 26

a, \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2.13=26\)

vậy............

b, \(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)

\(\Rightarrow x^2+2xy+y^2=a^2\)

\(\Rightarrow xy=\frac{a^2-b}{2}\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)

Vậy....

-x²-y²+8x+4y-21
=(-x²+8x-16)+(-y²+4y-4)-1
=-(x-4)²-(y-2)²-1
rồi đấy tự làm đi

bạn xài cái này gõ công thức ra đi

giúp man luôn nè : \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left(x-1\right)^3-4x.\left(x+1\right).\left(x-1\right)+3.\left(x-1\right).\left(x^2+x+1\right)\)

\(A=\left(x-1\right)^3-4x.\left(x^2-1^2\right)+3.\left(x^3-1\right)\)

Thay x=2 vào biểu thức ta có

\(A=\left(-2-1\right)^3-4.\left(-2\right).\left[\left(-2\right)^2-1\right]+3.\left[\left(-2\right)^3-1\right]\)

\(A=\left(-3\right)^3+8.3+3.\left(-9\right)\)

\(A=-27+24-27\)

\(A=-30\)