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a) \(x^2+2xy+y^2+x+y-2\le0\)
\(\Leftrightarrow\)\(\left(x+y\right)^2+x+y-2\le0\)
\(\Leftrightarrow\)\(\left(x+y+\frac{1}{2}\right)^2\le\frac{9}{4}\)
\(\Leftrightarrow\)\(-2\le x+y\le1\)
b) \(x^2+2y^2+2xy-16y-6x+30=0\)
\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)-6\left(x+y\right)=-y^2+10y-30\)
\(\Leftrightarrow\)\(\left(x+y\right)^2-6\left(x+y\right)=-\left(y^2-10y+25\right)-5\)
\(\Leftrightarrow\)\(\left(x+y-3\right)^2=-\left(y-5\right)^2+4\le4\)
\(\Leftrightarrow\)\(1\le x+y\le5\)
1)
a) \(2x^2-12x+18+2xy-6y\)
\(=2x^2-6x-6x+18+2xy-6y\)
\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)
\(=x\left(2y+2x-6\right)-3\left(2y+2x-6\right)\)
\(=\left(x-3\right)\left(2y+2x-6\right)\)
\(=2\left(x-3\right)\left(y+x-3\right)\)
b) \(x^2+4x-4y^2+8y\)
\(=x^2+4x-4y^2+8y+2xy-2xy\)
\(=\left(-4y^2+2xy+8y\right)+\left(-2xy+x^2+4x\right)\)
\(=2y\left(-2y+x+4\right)+x\left(-2y+x+4\right)\)
\(=\left(2y+x\right)\left(-2y+x+4\right)\)
2) \(5x^3-3x^2+10x-6=0\)
\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\Leftrightarrow\left(x^2+2\right)\left(5x-3\right)=0\)
Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow x^2+y^2-2x+4y+4+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
3)\(P\left(x\right)=x^2+y^2-2x+6y+12\)
\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)
\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)
Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
Bài làm
a) 2x2 - 12x + 18 + 2xy - 6y
= 2x2 - 6x - 6x + 18 + 2xy - 6y
= ( 2xy + 2x2 - 6x ) - ( 6y + 6x - 18 )
= 2x( y + x - 3 ) - 6( y + x - 3 )
= ( 2x - 6 ) ( y + x - 3 )
# Học tốt #
\(2x^2-2xy+2y^2-6x-6y+18=0\)
\(\Leftrightarrow x^2+x^2-2xy+y^2+y^2-6x-6y+9+9=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2=0\)
Mà \(\left(x-y\right)^2\ge0;\left(y-3\right)^2\ge0;\left(x-3\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}x-y=0\\y-3=0\\x-3=0\end{cases}}\Leftrightarrow z=y=3\)
b, x2 +y2+z2 +2x-4y-6z+14=0
<=> (x2+2x+1)+(y2-4y+4)+(z2-6z+9)=0
<=> (x+1)2+(y-2)2+(z-3)2=0
=>(x+1)2=(y-2)2=(z-3)2=0
=>x+1=y-2=z-3=0
=> x=-1; y=2; z=3
c, 2x2+y2-6x-4y+2xy+5=0
<=> (x2+y2+4+2xy-4x-4y)+(x2-2x+1)=0
<=> (x+y-2)2+(x-1)2=0
=> (x+y-2)2=(x-1)2=0
=>x+y-2=x-1=0
=>x=1; y=1
\(a)\) Có \(2012=x+y\ge2\sqrt{xy}\)\(\Leftrightarrow\)\(xy\le1006^2\)
\(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{x^2+2xy+y^2}+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\)
\(\le2+\frac{4.1006^2}{2012^2}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)
\(b)\) \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\ge\left(2+\frac{2012.4}{x+y}\right)^2\)
\(=\left(2+\frac{2012.4}{2012}\right)^2=36\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)
...
\(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)-2z\left(x-y\right)+z+\left(x^2-8x+16\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-y-z=0\\x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=1\\x=4\\y=3\end{cases}}\)
Vậy \(x=4\), \(y=3\), \(z=1\)
\(x^2-2xy+2y^2+2x-6y+1=0\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+1+\left(y^2-4x+4\right)-4=0\)
\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-2\right)^2-4=0\)
\(\Rightarrow\left(x-y+1\right)^2+\left(y-2\right)^2-4=0\)
Vì: \(\left(x-y+1\right)^2\ge0\), \(\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-y+1\right)^2+\left(y-2\right)^2-4\ge-4\)
Dấu ''='' xảy ra khi \(\hept{\begin{cases}x-y+1=0\\y-2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-y=-1\\y=2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-2=-1\\y=2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy GTNN của Biểu thức là -4 khi x=1 và y=2
pt \(\Leftrightarrow\)\(\left(\frac{1}{2}x^2-2xy+2y^2\right)+3\left(x-2x\right)=\frac{-1}{2}x^2+x-1\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(x-2y\right)^2+3\left(x-2y\right)+\frac{9}{2}=-\left(x-1\right)^2+9\)
\(\Leftrightarrow\)\(\left[\frac{1}{\sqrt{2}}\left(x-2y\right)+\frac{3}{\sqrt{2}}\right]^2=-\left(x-1\right)^2+9\le9\)
\(\Leftrightarrow\)\(-3\le\frac{1}{\sqrt{2}}\left(x-2y\right)+\frac{3}{\sqrt{2}}\le3\)
\(\Leftrightarrow\)\(-3\sqrt{2}-3\le x-2y\le3\sqrt{2}-3\)
\(\Leftrightarrow\)\(\frac{x-3\sqrt{2}+3}{2}\le y\le\frac{x+3\sqrt{2}+3}{2}\)
Với x càng lớn thì y càng lớn, x càng bé thì y càng bé => y ko có min, max