D = 2x^2 +y^2 - 2xy +2x -4y + 9
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20 tháng 2 2020
D=\(\left(x^2-2xy+y^2+4x-4y+4\right)+\left(x^2-2x+1\right)+4\)\(=\left(x+y+2\right)^2+\left(x+1\right)^2+4\ge4\Rightarrow Min_D=4\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x+y=-2\Rightarrow y=-1\end{matrix}\right.\)
TT
21 tháng 2 2020
Đặt \(A=2x^2+y^2-2xy+2x-4y+9\)
\(=\left(x^2-2xy+y^2\right)+2.\left(x-y\right).2+4+x^2-2x+5\)
\(=\left(x-y\right)^2+2.\left(x-y\right).2+2^2+\left(x^2-2x+1\right)+4\)
\(=\left(x-y+2\right)^2+\left(x-1\right)^2+4\)
Ta thấy : \(\hept{\begin{cases}\left(x-y+2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\forall x,y}\) \(\Rightarrow\left(x-y+2\right)^2+\left(x-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y+2\right)^2+\left(x-1\right)^2+4\ge4\forall x,y\)
hay : \(A\ge4\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}y=3\\x=1\end{cases}}\)
Vậy : min \(A=4\) tại \(\hept{\begin{cases}y=3\\x=1\end{cases}}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
31 tháng 8 2023
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
16 tháng 6 2018
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
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 \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
=(x-y)^2+x(x+2)-4y+9 câu trả lời đây nha 😊
⇒D=2².x²+y²-2xy+2x-4y+9
⇒D=4x²+y²-2xy-4y+9
⇒D=4²+2-4y+9
⇒D=16+2-4y+9
⇒D=18-36+9y
⇒D=-18+9y còn đây là cách làm nha