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10 tháng 8 2017
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
29 tháng 1 2019
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
KB
14 tháng 10 2018
Bài 1:
a ) \(Q=\dfrac{3}{2}x^2+x+1=\dfrac{3}{2}\left(x^2+\dfrac{2}{3}x+\dfrac{2}{3}\right)=\dfrac{3}{2}\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{5}{9}\right)=\dfrac{3}{2}\left[\left(x+\dfrac{1}{3}\right)^2+\dfrac{5}{9}\right]=\dfrac{3}{2}\left(x+\dfrac{1}{3}\right)^2+\dfrac{5}{6}\ge\dfrac{5}{6}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy Min Q là : \(\dfrac{5}{6}\Leftrightarrow x=-\dfrac{1}{3}\)
b ) \(R=x^2+2y^2+2xy-2y=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy Min R là : \(-1\Leftrightarrow x=-1;y=1\)
Bài 2 :
a ) \(Q=2x-2-3x^2\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{5}{9}\right)\)
\(=-3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{5}{9}\right]\)
\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\le-\dfrac{5}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy Max Q là : \(-\dfrac{5}{3}\Leftrightarrow x=\dfrac{1}{3}\)
b ) \(2-x^2-y^2-2\left(x+y\right)\)
\(=2-x^2-y^2-2x-2y\)
\(=-\left(x^2+2x+1\right)-\left(y^2+2y+1\right)+4\)
\(=-\left(x+1\right)^2-\left(y+1\right)^2+4\le4\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow x=y=-1\)
Vậy Max của b/t trên là : \(4\Leftrightarrow x=-1\)
c ) \(7-x^2-y^2-2\left(x+y\right)\)
\(=7-x^2-y^2-2x-2y\)
\(=-\left(x^2+2x+1\right)-\left(y^2+2y+1\right)+9\)
\(=-\left(x+1\right)^2-\left(y+1\right)^2+9\le9\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow x=y=-1\)
Vậy Max của b/t trên là : \(9\Leftrightarrow x=y=-1\)
25 tháng 8 2022
Bài 2:
a: \(=-\left(x^2+2x-100\right)\)
\(=-\left(x^2+2x+1-101\right)\)
\(=-\left(x+1\right)^2+101< =101\)
Dấu = xảy ra khi x=-1
b: \(=-3\left(x^2-\dfrac{1}{3}x\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}\right)\)
\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}< =\dfrac{1}{12}\)
Dấu = xảy ra khi x=1/6
c: \(=-\left(3x^2+4y^2-18x+8y-12\right)\)
\(=-\left(3x^2-18x+27+4y^2+8y+4-43\right)\)
\(=-3\left(x-3\right)^2-4\left(y+1\right)^2+43< =43\)
Dấu = xảy ra khi x=3 và y=-1
8 tháng 8 2020
a) x - 2y + x2 - 4y2
= x - 2y + ( x - 2y).( x+2y)
= (x - 2y).(1+ x + 2y)
b) x2- 4x2y2 + y2 + 2xy
= (x + y)2 - 4x2y2
= (x + y -2xy).(x + y + 2xy)
c) x6 - x4 + 2x3 + 2x2
= x2.(x4 - x2 + 2x + 2)
= x2.[x2.(x2 - 1) + 2(x+1)]
= x2.[x2.(x-1).(x+1) + 2(x+1)]
= x2.(x+1).(x2.(x-1)+2)
= x2.(x+1).(x3 - x2 + 2)
= x2.(x+1).(x3+x2-2x2+2)
= x2.(x+1).[x2( x+1) +2 (-x2 +1)]
= x2.(x+1).[x2( x+1) +2 (1+x).(1-x)]
= x2.(x+1)2.(x2 + 2 -2x)
d) x3 + 3x2 + 3x + 1- 8y3
= (x+1)3 - 8y3
= ( x+1 - 2y).[(x+1)2 + (x+1). 2y + 4y2]
= ( x + 1 - 2y).(x2 + 2x + 1 + 2xy + 2y + 4y2)
TG
8 tháng 8 2020
a/ \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+x^2-\left(2y\right)^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left[1+\left(x+2y\right)\right]\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b/ \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c/ \(x^6-x^4+2x^3+2x^2\)
\(=x^4\left(x^2-1\right)+2x^2\left(x+1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)+2x^2\left(x+1\right)\)
\(=\left(x+1\right)\left[x^4\left(x-1\right)+2x^2\right]\)
d/ \(x^3+3x^2+3x+1-8y^3\)
\(=\left(x+1\right)^3-\left(2y\right)^3\)
\(=\left(x+1-2y\right)\left[\left(x+1\right)^2+\left(x+1\right)2y+\left(2y\right)^2\right]\)
\(=\left(x+1-2y\right)\left[\left(x^2+2x+1\right)+2xy+2y+4y^2\right]\)
\(=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
`A= x^2+2xy-3x^2 +2y^2+3x^2-y^2`
`= (x^2-3x^2 +3x^2) +2xy +(2y^2 -y^2)`
`= x^2 +2xy +y^2`
`=(x+y)^2`
A = \(x^2\) + 2\(xy\) - 3\(x^2\) + 2y2 + 3\(x^2\) - y2
A = (\(x^2\)- 3\(x^2\) + 3\(x^2\)) + 2\(xy\) + (2\(y^2\) - y2)
A = \(x^2\) + 2\(xy\) + y2
A = (\(x\) + y)2