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`B = x^2- 2xy + y^2 + 2x - 10y + 17
`2B = 2x^2 - 4xy + 2y^2 + 4x - 20y + 34`
`= (x-y)^2 + (x+2)^2 + (y-5)^2 + 5 >= 5`.
Học tốt!
`2(x^2+y^2)+z^2=-2xy+2yz-4x-4`
`<=>2x^2+2y^2+z^2+2xy-2yz+4x+4=0`
`<=>(x^2+2xy+y^2)+(y^2-2yz+z^2)+(x^2+4x+4)=0`
`<=>(x+y)^2+(y-z)^2+(x+2)^2=0`
Vì `VT>=0`
Nên dấu "=" xảy ra khi `x+y=0,y-z=0,x+2=0`
`<=>x=-y,y=z,x=-2`
`<=>x=-2,y=z=-x=2`
Vậy `(x,y,z)=(-2,2,2)`
\(Q=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2x^2+2y^2}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2+2y^2+4xy}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}\)
$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$
$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$
$\Leftrightarrow x=3; y=-2$
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$B=9x^2+y^2+2z^2-18x+4z-6y+30$
$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$
$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$
$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
Vậy $B_{\min}=10$. Giá trị này đạt tại $(x-1)^2=(y-3)^2=(z+1)^2$
$\Leftrightarrow x=1; y=3; z=-1$
$C=x^2+y^2+z^2-xy-yz-xz+3$
$2C=2x^2+2y^2+2z^2-2xy-2yz-2xz+6$
$=(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)+6$
$=(x-y)^2+(y-z)^2+(z-x)^2+6\geq 6$
$\Rightarrow C\geq 3$
Vậy $C_{\min}=3$. Giá trị này đạt tại $x-y=y-z=z-x=0$
$\Leftrihgtarrow x=y=z$
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$D=5x^2+2y^2+4xy-2x+4y+2021$
$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$
$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$
$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$
$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$
Vậy $D_{\min}=2016$ khi $x+y+1=x-1=0$
$\Leftrightarrow x=1; y=-2$
c) \(\left(x+\dfrac{y}{x}\right)^3\)
\(=\left(\dfrac{x^2}{x}+\dfrac{y}{x}\right)^3\)
\(=\left(\dfrac{x^2+y}{x}\right)^3\)
\(=\dfrac{x^6+3x^4y+3x^2y^3+y^3}{x^3}\)
f) \(\left(x-\dfrac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3\)
\(=x^3-\dfrac{3}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{8}\)
h) \(\left(x+\dfrac{y^2}{2}\right)^3\)
\(=\left(\dfrac{2x}{2}+\dfrac{y^2}{2}\right)^3\)
\(=\left(\dfrac{2x+y^2}{2}\right)^3\)
\(=\dfrac{8x^3+12x^2y^2+6xy^4+y^6}{8}\)
k) \(\left(x-\dfrac{1}{3}\right)^3\)
\(=x^3-3\cdot x^2\cdot\dfrac{1}{3}+3\cdot x\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3\)
\(=x^3-x^2+\dfrac{x}{3}-\dfrac{1}{27}\)
m) \(\left(x+\dfrac{y^2}{3}\right)^3\)
\(=\left(\dfrac{3x}{3}+\dfrac{y^2}{3}\right)^3\)
\(=\left(\dfrac{3x+y^2}{3}\right)^3\)
\(=\dfrac{27x^3+27x^2y^2+9xy^4+y^6}{27}\)
Q) \(2\left(x^2+\dfrac{1}{2}y\right)\left(2x^2-y\right)\)
\(=2\left(2x^4-x^2y+x^2y-\dfrac{1}{2}y^2\right)\)
\(=2\left(2x^4-\dfrac{1}{2}y^2\right)\)
\(=4x^4-y^2\)
\(a,=x^3-16x-x^2-1-x^2+1=x^3-2x^2-16x\\ b,=y^4-81-y^4+4=-77\\ d,=a^2+b^2+c^2+2ab-2bc-2ac+a^2-2ac+c^2-2ab-2ac\\ =2a^2+b^2+2c^2-2bc-6ac\)