6xy + 3 - 9y + 2x = 0
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24 tháng 11 2023
b: \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9y^2+18y+9+y^2-6y-6-2y-23=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}10y^2+10y-20=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y^2+y-2=0\\x=3y+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(y+2\right)\left(y-1\right)=0\\x=3y+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\in\left\{-2;1\right\}\\x=3y+3\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(-3;-2\right);\left(6;1\right)\right\}\)
a: \(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9y=4x-6\\3x^2+6xy-x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{9}x-\dfrac{2}{3}\\3x^2+6x\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)-x+3\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x^2+\dfrac{8}{3}x^2-4x-x+\dfrac{4}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{17}{3}x^2-\dfrac{11}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x^2-11x-6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(17x+6\right)=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}17x+6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\cdot1-\dfrac{2}{3}=\dfrac{4}{9}-\dfrac{2}{3}=-\dfrac{2}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{6}{17}\\y=\dfrac{4}{9}\cdot\dfrac{-6}{17}-\dfrac{2}{3}=\dfrac{-14}{17}\end{matrix}\right.\end{matrix}\right.\)
EC
0
LD
2
21 tháng 10 2018
\(A=2x^2-6xy+9y^2-12x+2017\)
\(A=x^2+x^2-6xy+\left(3y\right)^2-12x+2014\)
\(A=\left(x^2-2\cdot x\cdot6+6^2\right)+\left[\left(3y\right)^2-2\cdot3y\cdot x+x^2\right]+1978\)
\(A=\left(x-6\right)^2+\left(3y-x\right)^2+1978\ge1978>0\forall x;y\)
P.s: 1978 năm sinh me t :)
AH
Akai Haruma
Giáo viên
9 tháng 9 2018
Lời giải:
Ta có \(2x^2-6xy+9y^2-6x+9=0\)
\(\Leftrightarrow (x^2-6xy+9y^2)+(x^2-6x+9)=0\)
\(\Leftrightarrow (x-3y)^2+(x-3)^2=0\)
Vì \((x-3y)^2; (x-3)^2\geq 0, \forall x,y\in\mathbb{R}\), do đó để \((x-3y)^2+(x-3)^2=0\) thì \(\left\{\begin{matrix} (x-3y)^2=0\\ (x-3)^2=0\end{matrix}\right.\Leftrightarrow x=3; y=1\)
Vậy........
9 tháng 9 2018
2x2 - 6xy + 9y2 - 6x + 9 = 0
<=> ( x2 - 6xy + 9y2 ) + ( x2 - 6x + 9 ) = 0
<=> ( x - 3y )2 + ( x - 3 )2 = 0
<=> x = 3; y = 1
Vậy x = 3 và y = 1
11 tháng 12 2019
\(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)
\(=\left(2x+3y\right)\left(2x-3y\right)^2-\left(2x-3y\right)\left(2x+3y\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y\right)\left(2x-3y-2x-3y\right)\)
\(=-\left(2x-3y\right)\left(2x+3y\right)\cdot6y\)
FN
2
T
16 tháng 9 2018
\(\Leftrightarrow\hept{\begin{cases}2.\left(x^2-3xy+5\right)=0\\3.\left(3y^2-4x+12\right)+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x^2-3xy+5=0\\3y^2-4x+12=-\frac{2}{3}\end{cases}}\)\(\Rightarrow x^2-3xy+5+3y^2-4x+12=-\frac{2}{3}\)
\(\Leftrightarrow3.\left(y^2-2.\frac{1}{2}x.y+\frac{x^2}{4}\right)+\frac{1}{4}x^2-4x+17+\frac{2}{3}=0\)
\(\Leftrightarrow3.\left(y-\frac{x}{2}\right)^2+\frac{1}{4}.\left(x^2-16x\right)+\frac{53}{3}=0\)
\(\Leftrightarrow3.\left(y-\frac{x}{2}\right)^2+\frac{1}{4}.\left(x^2-2.8x+64\right)+\frac{5}{3}=0\)
\(\Leftrightarrow3.\left(y-\frac{x}{2}\right)^2+\frac{1}{4}.\left(x-8\right)^2+\frac{5}{3}=0\)
Vì \(3.\left(y-\frac{x}{2}\right)^2+\frac{1}{4}.\left(x-8\right)^2\ge0\)nên \(3.\left(y-\frac{x}{2}\right)^2+\frac{1}{4}.\left(x-8\right)^2+\frac{5}{3}>0\)
Vậy không có x,y thỏa mãn ????
FM
16 tháng 9 2018
\(\hept{\begin{cases}2x^2-6xy+10=0\\9y^2-12x+26=0\end{cases}}\)
Cộng hai phương trình trên, ta có :
\(2x^2-6xy+10+9y^2-12x+26=0\Leftrightarrow\left(x^2-12x+36\right)+\left(9y^2-6xy+x^2\right)=0\)\(\Leftrightarrow\left(x-6\right)^2+\left(3y-x\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}x-6=0\\3Y-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\end{cases}}}\)
Bạn tự KL nhé
NV
1
TN
9 tháng 6 2017
Ta có:2x2-6xy+9y2-6x+9=0<=>(x2-6xy+9y2)+(x2-6x+9)=0
<=>(x-3y)2+(x-3)2=0
Vì (x-3y)2\(\ge0\);(x-3)2\(\ge0\) nên (x-3y)2+(x-3)2\(\ge0\)
Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}x-3y=0\\x-3=0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=3y\\x=3\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy x=3;y=1
TC
0
DK
2
9 tháng 11 2017
a) \(2x^3+6xy-x^2z-3yz\)
= \(\left(2x^3+6xy\right)-\left(x^2z+3yz\right)\)
=\(2x\left(x^2+3y\right)-z\left(x^2+3y\right)\)
=\(\left(x^2+2y\right)\left(2x-z\right)\)
b)\(x^2-6xy+9y^2-49\)
=\(x^2-2.x.3y+\left(3y\right)^2-7^2\)
=\(\left(x-3y\right)^2-7^2\)
=\(\left(x-3y+7\right)\left(x-3y-7\right)\)
