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x2 + 2y2 + z2 - 2xy - 2y - 4z + 5 = 0
<=> ( x2 - 2xy + y2 ) + ( y2 - 2y + 1 ) + ( z2 - 4z + 4 ) = 0
<=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2 = 0
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2\(\ge\)0\(\forall\)x ; y ; z
Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )
Thay ( 1 ) vào A , ta được :
\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)
Vậy A = 2
Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)
\(B=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)
\(=\frac{x^2y-x^2z+zy^2-xy^2+z^2x-z^2y}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)
\(=\frac{\left(x^2y-z^2y\right)-\left(xy^2-zy^2\right)-\left(x^2z-z^2x\right)}{\left(x^2-y^2\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x+z\right)-y^2-xz\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(xy+zy-y^2-xz\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[\left(xy-y^2\right)-\left(xz-zy\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x-y\right)-z\left(x-y\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{x-z}{x+y}\)
\(A=\frac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
\(=\frac{x^2y-y^2+x^2-y+x^2y^2-1}{x^2y+y^2+x^2+y+x^2y^2+1}\)
\(=\frac{\left(x^2y+x^2\right)+\left(x^2y^2-y^2\right)-\left(y+1\right)}{\left(x^2y+x^2\right)+\left(x^2y^2+y^2\right)+\left(y+1\right)}\)
\(=\frac{x^2\left(y+1\right)+y^2\left(x^2-1\right)-\left(y+1\right)}{x^2\left(y+1\right)+y^2\left(x^2+1\right)+\left(y+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y+1\right)+y^2\left(x^2-1\right)}{\left(x^2+1\right)\left(y+1\right)+y^2\left(x^2+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y^2+y+1\right)}{\left(x^2+1\right)\left(y^2+y+1\right)}\)
\(=\frac{x^2-1}{x^2+1}\)
\(a,A=5x^2a-10xya+5y^2a\)
\(=5a\left(x^2-2xy+y^2\right)\)
\(=5a\left(x-y\right)^2\)
Thay x = 124; y=24;a=2 ta có
\(5.2\left(124-24\right)^2=10.100^2=100000\)
\(b,B=2x^2+2y^2-x^2z+z-y^2z-2\)
\(=2\left(x^2+y^2-1\right)-z\left(x^2+y^2-1\right)\)
\(=\left(x^2+y^2-1\right)\left(2-z\right)\)
Thay x = 1 ; y = 1; z= -1 ta có
\(\left(1^2+1^2-1\right)\left(2-\left(-1\right)\right)=\left(1+1-1\right)\left(2+1\right)=1.3=3\)
\(c,C=x^2-y^2+2y-1\)
\(=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
Thay x = 75; y = 26 ta có
\(\left(75-26+1\right)\left(75+26-1\right)=50.100=5000\)
c)\(x^3+3xy+y^3\)
\(=x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=\left(x^2-xy+y^2\right)+3xy\)
\(=x^2-xy+y^2+3xy\)
\(=x^2+2xy+y^2=\left(x+y\right)^2\)
\(=1^2=1\)
a: \(5x-20y=5\left(x-4y\right)\)
b: \(x^2+x^2y+x^2y^2=x^2\left(1+y+y^2\right)\)
c: \(x\left(x+y\right)-\left(5x+5y\right)=\left(x+y\right)\left(x-5\right)\)
d: \(5\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(y+5\right)\)
\(A=2xy-x^2-2yz+4y^2=2\cdot2\cdot\dfrac{1}{2}-2^2-2\cdot\dfrac{1}{2}\cdot\left(-1\right)+4\cdot\left(\dfrac{1}{2}\right)^2\)
\(=2-4+1+4\cdot\dfrac{1}{4}=-2+1+1=0\)
A = 2 (2 . 1/2 - 2) - 2 . 1/2 (-1 - 2 . 1/2)
A = 2 (1 - 2) - 1 (-1 - 1)
A = 2 . (-1) - 1 . (-2)
A = -2 + 2
A = 0
@Liz.Ald2094