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\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b)Ta có: x2y+xy2+x+y=2010
<=>xy.x+xy.y+x+y=2010
<=>11x+11y+x+y=2010
<=>12(x+y)=2010
<=>x+y=167,5
=>(x+y)2=28056,25
<=>x2+y2+2xy=28056,25
<=>x2+y2=28034,25
a,\(x^2+2xy+7x+7y+y^2+10=\left(x^2+2xy+y^2\right)+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b,\(x^2y+xy^2+x+y=2010\Rightarrow xy\left(x+y\right)+x+y=2010\)
\(\Rightarrow12\left(x+y\right)=2010\Rightarrow x+y=167,5\)
Ta có:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)
Trả lời :
Ta có :
\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
Hok tốt
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y+2\right)\left(x+y+5\right).\)
b) \(x^2y+xy^2+x+y=2010\)
\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)
\(\Leftrightarrow11\left(x+y\right)+1\left(x+y\right)=2010\)
\(\Leftrightarrow12\left(x+y\right)=2010\)
\(\Leftrightarrow x+y=\frac{335}{2}\)
\(\Leftrightarrow\left(x+y\right)^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+2xy+y^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2+22=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2=\frac{112137}{4}.\)
Vậy \(x^2+y^2=\frac{112137}{4}.\)
a, = [(x-2).(x+1)]^2+(x-2)^2
= (x-2)^2.(x+1)^2+(x-2)^2
= (x-2)^2.[(x+1)^2+1]
= (x-2)^2.(x^2+2x+2)
Tk mk nha
b) \(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)
\(=\left(2x+1\right)\left(3x^4+3x^3+3x^2+3x^3+3x^2+3x+x^2+x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\left(3x^2+3x+1\right)\)
a, Xét : 3 - E = 3x^3-3xy-3y^3-x^3-xy-y^2/x^2-xy+y^2
= 2x^2-4xy+2y^2/x^2-xy+y^2
= 2.(x^2-2xy+y^2)/x^2-xy+y^2
= 2.(x-y)^2/x^2-xy+y^2
>= 0 ( vì x^2-xy+y^2 > 0 )
Dấu "=" xảy ra <=> x-y=0 <=> x=y
Vậy ..........
b, Có : (x+1995)^2 = x^2+3990+1995^2 = (x^2-3990x+1995^2)+7980x
= (x-1995)^2 + 7980x >= 7980x
=> M < = x/7980x = 1/7980 ( vì x > 0 )
Dấu "=" xảy ra <=> x-1995=0 <=> x=1995
Vậy ...............
1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)
3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)
5) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)=\left(x-1\right)^2\left(x+1\right)\)
6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
a: \(=\left(x^2+2xy+y^2\right)+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y+5\right)\left(x+y+2\right)\)
b: \(x^2y+xy^2+x+y=2010\)
\(\Leftrightarrow xy\left(x+y\right)+x+y=2010\)
\(\Leftrightarrow\left(x+y\right)\left(xy+1\right)=2010\)
\(\Leftrightarrow x+y=167.5\)
\(x^2+y^2=\left(x+y\right)^2-2xy=167.5^2-22=28034.25\)