Tìm x,y:
\(4x^2-4x+9y^2-6y+2=0\)
Tìm x,y:
4x^2 - 4x + 9y^2 - 6y + 2 = 0
Ta sẽ tạo các tổng bình phương như sau:
\(PT\Leftrightarrow\left(4x^2-4x+1\right)+\left(9y^2-6y+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2=0\)(1)
Do \(\left(2x-1\right)^2\ge0;\left(3y-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2+\left(3y-1\right)^2\ge0\)(2)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x-1=0\\3y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}}\)
Chu Văn Long thiếu câu dấu = xảy ra trước cái từ 1 và 2
Tim x,y biet:
1)x^2-2x+5+y^2-4y=0
2)4x^2+y^2-20x+26-2y=0
3)x^2+4y^2+13-6x-8y=0
4)4x^2+4x-6y+9x^2+2=0
5)x^2+y^2+6x-10y+34=0
6)25x^2-10x+9y^2-12y+5=0
7)x^2+9y^2-10x-12y+29=0
89x^2+12x+4y62+8y+8=0
9)4x^2+9y^2+20x-6y+26=0
10)3x^2+3y^2+6x-12y+15=0
11)x^2+4y^2+4x-4y+5=0
12)4x^2-12x+y^2-4y+13=0
13)x^2+y^2+2x-6y+10=0
14)4x^2+9y^2-4x+6y+2=0
15)y^2+2y+5-12x+9x^2=0
16)x^2+26+6y+9y^2-10x=0
17)10-6x+12y+9x^2+4y^2=0
18)16x^2+5+8x-4y+y^2=0
19)x^2+9y^2+4x+6y+5=0
20)5+9x^2+9y^2+6y-12x=0
21)x^2+20+9y62+8x-12y=0
22)x^2=4y+4y^2+26-10x=0
23)4y^2+34-10x+12y+x^2=0
24)-10x+y^2-8y+x^2+41=0
25)x^2+9y^2-12y+29-10x=0
26)9x^2+4y^2+4y+5-12x=0
27)4y^2-12x+12y+9x^2=13=0
28)4x^2+25-12x-8y+y^2=0
29)x62+17+4y^2+8x+4y=0
30)4y^2+12y+25+8x+x^2=0
31)x^2+20+9y^2+8x-12y=0
giup mk voi minh can gap ak, cam on cac ban
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 biết x^2+y^2-4x+6y+13=0
(x2-4x+4) + (y2+6y+9) = 0
bạn làm tiếp nhé, dáp số x=2, y=-3
cho x+y+z thỏa mãn đẳng thức 4x^2 + 9y^2 + 16z^2 - 4x - 6y - 8z + 3 = 0
khi đó xy+yz+zx=
Tìm x,y bt:
a) x2 + y2 - 2x + 4y + 5 = 0
b) 4x2 + 9y2 - 4x - 6y + 2 = 0
a) x2 + y2 - 2x + 4y + 5 = 0
\(\Leftrightarrow\)( x2 - 2x + 1 ) + ( y2 + 4y + 4 ) = 0
\(\Leftrightarrow\)( x - 1 )2 + ( y + 2 )2 = 0
\(\Rightarrow\)x - 1 = 0 và y + 2 = 0
\(\Rightarrow\)x = 1 và y = - 2
Vậy : x = 1 và y = - 2
b) 4x2 + 9y2 - 4x - 6y + 2 = 0
\(\Leftrightarrow\)[ ( 2x )2 - 4x + 1 ] + [ ( 3y )2 - 6y + 1 ] = 0
\(\Leftrightarrow\)( 2x - 1 )2 + ( 3y - 1 )2 = 0
\(\Rightarrow\)2x - 1 = 0 và 3y - 1 = 0
\(\Rightarrow\)x = 1 / 2 và y = 1 / 3
Vậy : x = 1 / 2 và y = 1 / 3
a) \(x^2+y^2-2x+4y+5=0\)
\(x^2+y^2-2x+4y+1+4=0\)
\(\left(x^2-2x+1\right)\left(y^2+4y+4\right)=0\)
\(\left(x-1\right)^2\left(y+2\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\y=-2\end{cases}}\)
b) \(4x^2+9y^2-4x-6y+2=0\)
\(\left(4x^2-4x+1\right)\left(9y^2-6y+1\right)=0\)
\(\left(2x-1\right)^2\left(3y-1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\3y-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}}\)
Tìm x, y, z biết
4x2 + 9y2 + 16z2 - 4x - 6y - 8z + 3 = 0
\(4x^2-4x+1+9y^2-6y+1+16z^2-8z+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2+\left(4z-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\3y-1=0\\4z-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\\x=\frac{1}{4}\end{cases}}\)
vay ................................................
Ta có :
4x2 + 9y2 + 16z2 - 4x - 6y - 8z + 3 = 0
( 2x ) 2 + ( 3y)2 + ( 4z)2 - 4x - 6y - 8z + 3 = 0
\([\left(2x\right)^2-2.2x+1]+[\left(3y\right)^2-2.3y+1]+[\left(4z\right)^2-2.4z+1]=0\)=0
( 2x-1)2 + ( 3y -1 )2 + ( 4z - 1) 2 = 0
Mà ( 2x-1)2 \(\ge\)0 với mọi x
( 3y-1 )2 \(\ge0\)với mọi y
( 4z - 1) 2 \(\ge0\)với mọi z
nên \(\hept{\begin{cases}2x-1=0\\3y-1=0\\4z-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\\z=\frac{1}{4}\end{cases}}}\)
Vậy x = 1/2 ; y = 1/3 ; z = 1/4
Bài 4:
a, Tìm GTLN
\(Q=-x^2-y^2+4x-4y+2\)
b, Tìm GTLN
\(A=-x^2-6x+5\)
\(B=-4x^2-9y^2-4x+6y+3\)
c, TÌm GTNN
\(P=x^2+y^2-2x+6y+12\)
a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)
\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)
\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)
Vậy MaxQ=10 khi x=2, y=-2
b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)
\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)
Vậy MaxA=14 khi x=-3
+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)
Vậy MaxB=5 khi x=-1/2, y=1/3
c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)
Vậy MinP=2 khi x=1, y=-3