tim x y z biết
a,4x^2+9y^2+4x-24y+17=0
b,2x^2+2y^2+z^2+2xy-2xz-6y+9=0
c,x^2+2y+2xy+2x+6y+5=0
tim x y z biết
a,4x^2+9y^2+4x-24y+17=0
b,2x^2+2y^2+z^2+2xy-2xz-6y+9=0
c,x^2+2y+2xy+2x+6y+5=0
\(a,4x^2+9y^2+4x-24y+17=0\)
\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)
\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)
Tìm x,y,z biết: a) x^2+y^2-4x+4y+8=0 b) 5x^2-4xy+y^2=0 c) x^2+2y^2+z^2-2xy-2y-4z+5=0 d) 3x^2+3y^2+3xy-3x+3y+3=0 e) 2x^2+y^2+2z^2-2xy-2xz+2yz-2z-2z-2x+2=0
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
d)3x2+3y2+3xy-3x+3y+3=0
⇔ 6x2+6y2+6xy-6x+6y+6=0
⇔ 3(x+y)2+3(x-1)2+3(y+1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
tìm x,y biết:
1) 5x2 + 3y2 + z2 - 4z + 6xy + 4z + 6 = 0
2) 2x2 + 2y2 + z2 + 2xy + 2xz + 2x + 4y + 5 = 0
3) 2x2 + 2y2 + z2 + 2xy +2xz + 2yz + 10x + 6y + 34 = 0
Tìm x, y thuộc Z để:
a) xy + x - y = 2
b) x - 2xy + y = 0
c) x. (x - 2) - (2 - x)y - 2. (x - 2) = 3
d) (2x - y). (4x2 + 2xy + y2) + (2x + y). (4x2 - 2xy + y2) - 16x. (x2 - y) = 32
e) x2 - 2xy + 2y2 - 2x + 6y +5 = 0
g) x2 + 2xy + 7x + 7y + 2y2 = 0
tìm x,y,z biết:2x2+2y2+z2+2xy+2xz+2yz+10x+6y+34=0
2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0
(x2 + y2 + z2 + 2xy + 2xz + 2yz) + (x2 + 10x + 25) + (y2+ 6y + 9) = 0
( x + y + z)2 + ( x + 5)2 + (y + 3)2 = 0
( x + y + z)2 = 0 ;
( x + 5)2 = 0 ;
(y + 3)2 = 0
vậy x = - 5 ; y = -3; z = 8
Tìm x, y, z biết rằng: 2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0
Giải
2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0
(x2 + y2 + z2 + 2xy + 2xz + 2yz) + (x2 + 10x + 25) + (y2+ 6y + 9) = 0
( x + y + z)2 + ( x + 5)2 + (y + 3)2 = 0
( x + y + z)2 = 0 ; ( x + 5)2 = 0 ; (y + 3)2 = 0
x = - 5 ; y = -3; z = 8
Phần trả lời của mình cũng giống như Arcobaleno vậy
tìm x,y,z biết
2x^2 + 2y^2 +z^2 + 2xy + 2xz + 2yz + 10x + 6y + 34=0
tìm gtnn
A= 2x^2 + 4y^2 +4xy + 2x + 4y +9
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
\(A=2x^2+4y^2+4xy+2x+4y+9=\left(x^2+4y^2+4xy+2x+4y+1\right)+x^2+8\)
\(=\left(x+2y+1\right)^2+x^2+8\ge8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}}\)
Vậy \(Min\left(A\right)=8\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}\)
Tìm x; y; z biết:
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(2x^2+2y^2+z^2+2xy+2yz+2xz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
\(\Rightarrow x=-5,y=-3,z=8\)
Cho 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz +2xz +10x +6y +34 =0 Tìm x, y,z
Ta có:
\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2+2zx+x^2\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)+z^2=0\)\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2+\left(x+5\right)^2+\left(y+3\right)^2+z^2=0\)
Không tồn tại x,y,z thỏa mãn đề bài
Tìm x; y; z biết:
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}z=8\\x=-5\\y=-3\end{matrix}\right.\)
Vậy x = -5; y = -3; z = 8