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\(x^2-2xy+2y^2+2x-6y+1=0\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+1+\left(y^2-4x+4\right)-4=0\)
\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-2\right)^2-4=0\)
\(\Rightarrow\left(x-y+1\right)^2+\left(y-2\right)^2-4=0\)
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
\(a)\) 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 \(\Leftrightarrow\)\(x=y=1006\)
...
a) \(x^2+4y^2-6x-4y+10=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-4y+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\2y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)
b) \(2x^2+y^2+2xy-10x+25=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x-5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-5\\x=5\end{cases}}\)
c) \(x^2+2xy+4x-4y-2xy+5=0\)
\(\Leftrightarrow x^2-4x-4y+5=0\)
Xem lại đề câu c).
a) x2 + 4y2 - 6x - 4y + 10 = 0
<=> x2 - 6x + 9 + 4y2 - 4y + 1 = 0
<=> ( x - 3 )2 + ( 4y - 1 )2 = 0
<=> \(\hept{\begin{cases}x-3=0\\4y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{4}\end{cases}}\)
b) 2x2 + y2 + 2xy - 10x + 25 = 0
<=> x2 + 2xy + y2 + x2 - 10x + 25 = 0
<=> ( x + y )2 + ( x - 5 )2 = 0
<=> \(\hept{\begin{cases}x+y=0\\x-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5\\x=5\end{cases}}\)
c) Xem lại đề
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
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
Cho \(x\)và \(y\)thỏa mãn \(x^2\)+ \(2xy+6x+6y+2y^2+8=0\)
Tìm GTLN. GTNN của biểu thức \(B=x+y+2010\)
pt \(\Leftrightarrow\)\(\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}=-y^2+\frac{49}{4}-10\)
\(\Leftrightarrow\)\(\left(x+y+\frac{7}{2}\right)^2=-y^2+\frac{9}{4}\le\frac{9}{4}\)
\(\Leftrightarrow\)\(\frac{-3}{2}\le x+y+\frac{7}{2}\le\frac{3}{2}\)
\(\Leftrightarrow\)\(-4\le x+y+1\le-1\)
Dấu "=" tự xét nhé
\(x^2+2y^2+2xy+3x+3y-4=0\)
<=> \(x^2+2xy+y^2+3\left(x+y\right)+y^2-4=0\)
<=> \(\left(x+y\right)^2+3\left(x+y\right)-4+y^2=0\)
<=>\(A^2+3A-4+y^2=0\)
<=> (A-1)(A+4)=-y2\(\le0\)
do A-1 <A+4
=> \(\left\{{}\begin{matrix}A-1\le0\\A+4\ge4\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}A\le1\\A\ge-4\end{matrix}\right.\)
<=> \(-4\le A\le1\)
minA xảy ra <=> \(\left\{{}\begin{matrix}y=0\\x+y=-4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=0\\x=-4\end{matrix}\right.\)(t/m)
maxA xảy ra <=> \(\left\{{}\begin{matrix}y=0\\x+y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=0\\x=1\end{matrix}\right.\)(t/m)
Vũ Minh TuấnTrần Thanh PhươngLê Thị Thục HiềnBăng Băng 2k6 giúp vs
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) \(x^2+2y^2+2xy-16y-6x+30=0\)
\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)-6\left(x+y\right)=-y^2+10y-30\)
\(\Leftrightarrow\)\(\left(x+y\right)^2-6\left(x+y\right)=-\left(y^2-10y+25\right)-5\)
\(\Leftrightarrow\)\(\left(x+y-3\right)^2=-\left(y-5\right)^2+4\le4\)
\(\Leftrightarrow\)\(1\le x+y\le5\)