tìm x,y
4x2-4x+9y2-6y+2=0
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29 tháng 6 2021
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
29 tháng 6 2021
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
11 tháng 10 2021
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+2\right)\)
AC
2
27 tháng 8 2021
\(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(x^2-3y^2=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\)
\(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]=\left(3x-3y-2x+2y\right)\left(3x-3y+2x+2y\right)=\left(x-y\right)\left(5x-y\right)\)
\(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
\(27x^3-0,001=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
\(125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
27 tháng 8 2021
a: \(x^4-y^4=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
c: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
d: \(\left(4x^2-4x+1\right)-\left(x+1\right)^2=\left(2x-1\right)^2-\left(x+1\right)^2\)
\(=\left(2x-1-x-1\right)\left(2x-1+x+1\right)\)
\(=3x\left(x-2\right)\)
e: \(x^3+27=\left(x+3\right)\left(x^2+3x+9\right)\)
18 tháng 12 2020
x2+y2-4x+6y+13=0
(x2-4x+4)+(y2+6y+9)=0
(x-2)2+(y+3)2=0
suy ra x-2=0 hoặc y+3=0
*x-2=0=>x=2 *y+3 =0=> y=-3
vậy x=2,y=-3
HP
1
6 tháng 9 2020
x2 + 2y2 + 2xy - 4x + 6y + 29 = 0
<=> ( x2 + 2xy + y2 - 4x - 4y + 4 ) + ( y2 + 10y + 25 ) = 0
<=> [ ( x2 + 2xy + y2 ) - 2( x + y ).2 + 22 ] + ( y + 5 )2 = 0
<=> ( x + y - 2 )2 + ( y + 5 )2 = 0 (*)
<=> \(\hept{\begin{cases}\left(x+y-2\right)^2\ge0\forall x,y\\\left(y+5\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+y-2\right)^2+\left(y+5\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x+y-2=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\\y=-5\end{cases}}\)
Vậy x = 7 ; y = -5
\(4x^2-4x+1+9y^2-6y+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Ta có:4x2-4x+9y2-6y+2=0
<=>(4x2-4x+1)+(9y2-6y+1)=0
<=> (2x-1)2+(3y-1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)