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KQ
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30 tháng 9 2019
a) \(x^2+7x+7y-y^2\)
\(=\left(x+y\right)\left(x-y\right)+7\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+7\right)\)
b) \(x^2-xy-6y^2\)
\(=-6y^2-3xy+2xy+x^2\)
\(=-3y\left(2y+x\right)+x\left(2y+x\right)\)
\(=\left(x-3y\right)\left(2y+x\right)\)
c) \(x^2-3x^2-6x+8\)
\(=\left(x+2\right)\left(x^2-2x+4\right)-3x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-5x+4\right)\)
\(=\left(x+2\right)\left(x^2-4x-x+4\right)\)
\(=\left(x+2\right)\left[x\left(x-4\right)-\left(x-4\right)\right]\)
\(=\left(x+2\right)\left(x-1\right)\left(x-4\right)\)
30 tháng 9 2019
a)x2+7x+7y-y2=(x-y)(x+y)+7.(x+y)
=(x+y)(x-y+7)
b)x2-xy-6y2=x2-xy-4y2-2y2
=(x-2y)(x+2y)-y(x-2y)
=(x-2y)(x+2y-y)
KT
4 tháng 2 2018
\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
4 tháng 2 2018
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
BN
10 tháng 2 2019
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
10 tháng 2 2019
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 \(x^2+y^2=\frac{112137}{4}.\)
VV
16 tháng 6 2017
1/ \(2x^2+5xy+y^2=2\left(x^2+2x.\frac{5y}{4}+\frac{25y^2}{16}-\frac{17y^2}{16}\right)\)
\(=2\left[\left(x+\frac{5y}{4}\right)^2-\left(\frac{\sqrt{17}y}{4}\right)^2\right]=2\left(x+\frac{5-\sqrt{17}}{4}y\right)\left(x+\frac{5+\sqrt{17}}{4}y\right)\)
2/ \(x^2-3xy-2y^2=x^2-2x.\frac{3y}{2}+\frac{9y^2}{4}-\frac{17y^2}{4}\)
\(=\left(x-\frac{3y}{2}\right)^2-\left(\frac{\sqrt{17}y}{2}\right)^2=\left(x+\frac{-3-\sqrt{17}}{2}y\right)\left(x+\frac{-3+\sqrt{17}}{2}y\right)\)
3/ \(6x^2+xy-7x-2y^2+7y-5\)
\(=6x^2+4xy-10x-3xy-2y^2+5y+3x+2y-5\)
\(=2x\left(3x+2y-5\right)-y\left(3x+2y-5\right)+3x+2y-5\)
\(=\left(3x+2y-5\right)\left(2x-y+1\right)\)
4/ \(6a^2-ab-2b^2+a+4b-2\)
\(=6a^2-4ab+4a+3ab-2b^2+2b-3a+2b-2\)
\(=2a\left(3a-2b+2\right)+b\left(3a-2b+2\right)-\left(3a-2b+2\right)\)
\(=\left(3a-2b+2\right)\left(2a+b-1\right)\)
30 tháng 1 2019
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:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)