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KQ
1
S
12 tháng 8 2018
<=>(x2+y2+z2+2xy+2yz+2xz)+(x2+2x+1)+(y2+4y+4)=0
<=>(x+y+z)2+(x+1)2+(y+2)2=0
Mà \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\Rightarrow\left(x+y+z\right)^2+\left(x+1\right)^2+\left(y+2\right)^2\ge0}\)
=>\(\hept{\begin{cases}x+y+z=0\\x+1=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}z=3\\x=-1\\y=-2\end{cases}}}\)
KN
10 tháng 9 2019
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 \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
10 tháng 9 2019
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 #
ND
27 tháng 6 2018
a. \(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
O
2
TT
6 tháng 8 2020
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).
6 tháng 8 2020
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
11 tháng 6 2016
a) Ta có \(x^2+y^2+2x-4y+5=0\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)
<=> x=-1;y=2
b)Ta có:\(x^2+4y^2-x+4y+\frac{5}{4}=0\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+1\right)=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)
<=> x=1/2 ;y=-1/2
HT
11 tháng 6 2016
a, \(x^2+y^2+2x-4y+5=0\Rightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0.\)
\(\left(x+1\right)^2+\left(y-2\right)^2=0\)
\(\Rightarrow x+1=0\)và \(y-2=0\)
\(\left(+\right)x+1=0\Rightarrow x=-1\)
\(\left(+\right)y-2=0\Rightarrow y=2\)
Vậy x=-1 ; y=2
b, \(x^2+4y^2-x+4y+\frac{5}{4}=0\)
\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+\frac{4}{4}\right)=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\) và \(2y+1=0\)
\(\left(+\right)x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
\(\left(+\right)2y+1=0\Rightarrow2y=-1\Rightarrow y=-\frac{1}{2}\)
Vậy \(x=\frac{1}{2};y=-\frac{1}{2}\)
DA
2
8 tháng 10 2018
\(x^2+y^2-2x+4y+5=0\)
\(\left(x^2-2x+1\right)+\left(y^2+2.y.2+2^2\right)=0\)
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x;y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Vậy \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
9 tháng 12 2018
1
a) x2 + 4y2 + 4xy - 16
=(x2 + 4xy + 4y2) - 16
=(x+2y)2 - 16
=(x+2y-4)(x+2y+4)
b)x2 + y2 - 2x + 4y + 5 =0
<=> x2 - 2x + 1 + y2 - 4y + 4=0
<=> (x-1)2 + (y-2)2 =0
<=> x=1 và y=2
18 tháng 9 2018
a) (x2+2x+1)+(y2+2y+1)=0
=>(x+1)2+(y+1)2=0
Vì\(\left(x+1\right)^2\ge0;\left(y+1\right)^2\ge0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
Vậy x=y=-1
Bạn làm tiếp câu còn lại nha <3
Chúc bạn học tốt :)
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
Ta thấy: \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Do đó: \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}\). Vậy ...