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rút gọn P=2/x-(x2/(x2-xy)+(x2-y2)/xy-y2/(y2-xy)):(x2-xy+y2)/(x-y)
r tìm gt P với |2x-1|=1 ; |y+1|=1/2
\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x^2+y^2+2xy\right)+\left(x^2y^2+2xy+1\right)+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y\right)^2+\left(1+xy\right)^2+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y+1+xy\right)^2\) là SCP
(1+x2)(1+y2)+4xy+2(x+y)(1+xy)
= 1+y2+x2+x2y2+2xy+2xy+2(x+y)(1+xy)
=(x2+2xy+y2)+(x2y2+2xy+1)+2(x+y)(1+xy)
=(x+y)2+(xy+1)2+2(x+y)(1+xy)
=(x+y+xy+1)2
\(A=x^2+y^2+xy-3x-3y+2-18\)
\(=\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)+2015\)\(=\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+2015\ge2015\)
\(A_{min}=2015\) khi \(\left(x;y\right)=\left(1;1\right)\)
a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\)
\(=\left(x^2+y^2\right)\left(x^3+y^3\right)-\left(xy\right)^2\left(x+y\right)\)
\(=10.26-\left(-3\right)^2.2=...\)
(x+y)5=32
⇔ x5+5x4y+10x3y2+10x2y3+5xy4+y5 = 32
⇔ x5+y5 = 32-5xy(x3+y3)-10x2y2(x+y)
= 32-5.(-3).26-10.(-3)2.2
= 242
Ta có :
2(xy - x^2 - y + 1008) = y^2 + 2018
<=> 2xy - 2x^2 - 2y + 2016 = y^2 + 2018
<=> 2xy - 2x^2 - 2y = y^2 + 2
<=> 2xy - 2x^2 - 2y - y^2 - 2 = 0
<=> -(2x^2 - 2xy + y^2/2) - y^2/2 - 2y - 2 = 0
<=> -2(x^2 - xy + y^2/4) - 2(y^2/4 + y + 1) = 0
<=> -2(x-y/2)^2 - 2(y/2 + 1)^2 = 0
<=> 2(x-y/2)^2 + 2(y/2 + 1)^2 = 0
Dấu " = " xảy ra <=> x - y/2 = 0 ; y/2 + 1 = 0
<=> x = y/2 ; y = -2
<=> x = -1 ; y = -2
Vậy x = -1 ; y = -2
Ta có :
2(xy - x^2 - y + 1008) = y^2 + 2018
<=> 2xy - 2x^2 - 2y + 2016 = y^2 + 2018
<=> 2xy - 2x^2 - 2y = y^2 + 2
<=> 2xy - 2x^2 - 2y - y^2 - 2 = 0
<=> -(2x^2 - 2xy + y^2/2) - y^2/2 - 2y - 2 = 0
<=> -2(x^2 - xy + y^2/4) - 2(y^2/4 + y + 1) = 0
<=> -2(x-y/2)^2 - 2(y/2 + 1)^2 = 0
<=> 2(x-y/2)^2 + 2(y/2 + 1)^2 = 0
Dấu " = " xảy ra <=> x - y/2 = 0 ; y/2 + 1 = 0
<=> x = y/2 ; y = -2
<=> x = -1 ; y = -2
Vậy x = -1 ; y = -2