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ML
3
LT
31 tháng 7 2017
2C=4x^2+2x-10=((2x)^2+4x\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\))-\(\dfrac{41}{4}\)
=\(\left(2x+\dfrac{1}{2}\right)^2\)-41/4\(\ge\dfrac{-41}{4}\)
=> C\(\ge\dfrac{-41}{8}\)
Vậy min C = \(\dfrac{-41}{8}\)khi x=\(\dfrac{-1}{4}\)
LT
31 tháng 7 2017
\(4A=4x^2+44y^2+24xy-8y+20=\left(2x\right)^2+2.2x.6y+\left(6y\right)^2+8y^2-8y+20=\left(2x+6y\right)^2+2\left(4y^2-4y+1\right)+18=\left(2x+6y\right)^2+2\left(2y-1\right)^2+18\ge18\)
TC
27 tháng 6 2016
a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
7 tháng 8 2017
Bài 2:
a, \(x^2-6x+10=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\ge1>0\)
\(\Rightarrowđpcm\)
b, \(x^2-4xy+4y^2+1=\left(x-2y\right)^2+1>0\)
\(\Rightarrowđpcm\)
c, \(x^2-4x+7=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\ge3\)
\(\Rightarrowđpcm\)
d, \(x^2+y^2-2x+4y+5\)
\(=x^2-2x+1+y^2+4y+4\)
\(=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
\(\Rightarrowđpcm\)
DH
10 tháng 8 2017
Ép người quá đáng >.<
Bài 1:
a, \(-\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)+\left(2x^2+1\right)\)
\(=-\left(4x^4-4x^3+2x^2+4x^3-4x^2+2x+2x^2-2x+1\right)+2x^2+1\)
\(=-\left(4x^4+1\right)+2x^2+1=-4x^4+2x^2\)
b, \(\left(x^2+x+2\right)^2+\left(x-1\right)^2-2\left(x^2+x+2\right)\left(x-1\right)\)
\(=\left(x^2+x+2-x+1\right)^2=\left(x^2+3\right)^2\)
d, \(-125x^3+225x^2-135x+27\)
\(=-\left(125x^3-225x^2+135x-27\right)\)
\(=-\left(125x^3-75x^2-150x^2+90x+45x-27\right)\)
\(=-\left[25x^2\left(5x-3\right)-30x\left(5x-3\right)+9\left(5x-3\right)\right]\)
\(=-\left[\left(5x-3\right)\left(25x^2-15x-15x+9\right)\right]\)
\(=-\left(5x-3\right)^3\)
CQ
3
23 tháng 11 2017
\(Câu\text{ }1:\\ A=-2x^2-y^2-2xy+4x+2y+5\\ =-x^2-x^2-y^2-2xy+2x+2x+2y-1-1+7\\ =-\left(x^2+2xy+y^2\right)+\left(2x+2y\right)-1-\left(x^2-2x+1\right)+7\\ =-\left(x+y\right)^2+2\left(x+y\right)-1-\left(x-1\right)^2+7\\ =-\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-\left(x-1\right)^2+7\\ =-\left(x+y-1\right)^2-\left(x-1\right)^2+7\\ =-\left[\left(x+y-1\right)^2+\left(x-1\right)^2\right]+7\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \left(x+y-1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-1\right)^2+\left(x+y-1\right)^2\ge0\forall x;y\\ \Rightarrow-\left[\left(x-1\right)^2+\left(x+y-1\right)^2\right]\le0\forall x;y\\ \Rightarrow A=-\left[\left(x-1\right)^2+\left(x+y-1\right)^2\right]+7\le7\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+y-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y+1-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\\ Vậy\text{ }A_{\left(Max\right)}=7\text{ }khi\text{ }\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
\(Câu\text{ }2:\\ B=2x^2+4y^2+4xy+2x+4y+9\\ =x^2+x^2+4y^2+4xy+2x+4y+1+8\\ =\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+x^2+1+8\\ =\left(x+2y\right)^2+2\left(x+2y\right)+1+x^2+8\\=\left[\left(x+2y\right)^2+2\left(x+2y\right)+1\right]+x^2+8\\ =\left(x+2y+1\right)^2+x^2+8\\ Do\text{ }x^2\ge0\forall x\\ \left(x+2y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2+8\ge8\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}x^2=0\\\left(x+2y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\\ Vậy\text{ }B_{\left(Min\right)}=8\text{ }khi\text{ }\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right. \)
\(\)
8 tháng 11 2018
a) \(4x\left(x-5\right)+3y\left(x-5\right)\)
\(=\left(x-5\right)\left(4x+3y\right)\)
b) \(x^2-2x-4y^2-4y\)
\(=\left[x^2-\left(2y\right)^2\right]-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(x^2+x-y^2+y\)
\(=\left(x^2-y^2\right)+\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+1\right)\)
d) \(3x^2+3y^2-6xy-12\)
\(=3\left(x^2+y^2-2xy-4\right)\)
\(=3\left[\left(x-y\right)^2-2^2\right]\)
\(=3\left(x-y-2\right)\left(x-y+2\right)\)
30 tháng 1 2022
\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)
\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)
\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)
LK
29 tháng 6 2018
P = 3x2 - 2x + 3y2 - 2y + 6xy +2018
P = 3(x2 + y2 + 2xy) - 2(x + y) + 2018
P = 3[(x + y)2 - 2xy + 2xy] -2.5 + 2018
P = 3[ 52 +0] - 10 + 2018
P = 3.25 + 2008
P = 75 + 2008
P = 2083
18 tháng 9 2018
a) (x2+2x+1)+(y2+2y+1)=0
=>(x+1)2+(y+1)2=0
Vì\(\left(x+1\right)^2\ge0;\left(y+1\right)^2\ge0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
Vậy x=y=-1
Bạn làm tiếp câu còn lại nha <3
Chúc bạn học tốt :)
Đề là gì ạ ?
tim MIN cua 2 da thuc