Tìm GTNN của:
A=x2+y2+xy-x+y+1
Biết xy=11 và x2y+xy2+x+y=2010. Tính x2+y2.
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a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)
Lời giải:
a.
\(\frac{x^2+xy+x+y}{x^2-xy+x-y}=\frac{x(x+y)+(x+y)}{x(x+1)-y(x+1)}=\frac{(x+y)(x+1)}{(x+1)(x-y)}=\frac{x+y}{x-y}\)
b.
\(\frac{x^2-6x+9}{3x^2-9x}=\frac{(x-3)^2}{3x(x-3)}=\frac{x-3}{3x}\)
c.
\(\frac{y^2-x^2}{x^2y-xy^2}=\frac{(y-x)(y+x)}{-xy(y-x)}=\frac{x+y}{-xy}\)
\(a,=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\\ b,=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\\ c,=x\left(x+1\right)+a\left(x+1\right)=\left(x+a\right)\left(x+1\right)\\ d,Sửa:x^2y+xy^2-3x-3y=xy\left(x+y\right)-3\left(x+y\right)=\left(xy-3\right)\left(x+y\right)\\ e,=xy\left(x+1\right)-\left(x+1\right)=\left(xy-1\right)\left(x+1\right)\\ f,=x^2-4=\left(x-2\right)\left(x+2\right)\\ g,=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\\ h,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ i,=\left(x-4\right)^2-24y^2=\left(x-2\sqrt{6}y-4\right)\left(x+2\sqrt{6}y+4\right)\)
a) \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+\left(xy-2y\right)\)
\(=4\left(x-2\right)\left(x-2\right)+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8\right)+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8+x-2\right)\)
\(=\left(x-2\right)\left(5x-10\right)\)
\(=5\left(x-2\right)^2\)
a, \(=4\left(x-2\right)^2+y\left(x-2\right)=\left(x-2\right)\left(4x-8+y\right)\)
b, \(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-xy+y^2-y^2\right]=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)=x\left(x-y\right)\left(x^2-2xy+y^2-y\right)\)
c, \(=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\)
d, không phân tích được
a) \(3x\left(5x^2-2x-1\right)\)
\(=3x.5x^2-3x.2x+3x.\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^3-2xy+3\right)\left(-xy\right)\)
\(=\left(-xy\right).\left(x^2+2xy-3\right)\)
\(=\left(-xy\right).x^2+\left(-xy\right).2xy+\left(-xy\right).\left(-3\right)\)
\(=x^3y-2x^2y^2+3xy\)
mấy câu sau vt lại đè
c)x2y(2x3 - xy2 - 1);
d)x(1,4x - 3,5y);
e)xy(x2 - xy + y2);
f)(1 + 2x - x2)5x;
g) (x2y - xy + xy2 + y3). 3xy2;
h) x2y(15x - 0,9y + 6);
Đây ạ giúp mik vs bt tết đs mng :<
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
Sửa đề
\(2A=2x^2+2y^2+2xy-2x+2y+2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)\)
\(=\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
\(\Rightarrow A_{min}=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(x^2y+xy^2+x+y=xy\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(xy+1\right)=12\left(x+y\right)=2010\)
\(\Rightarrow x+y=\dfrac{2010}{12}\)
\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy=\left(\dfrac{2010}{12}\right)^2-2\cdot11=\dfrac{112137}{4}\)