Viết dưới dạng tổng 2 bình phương
x2-2xy+2y2+2y+1
x2+10x+26+y2+2y
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Lời giải:
a. $x^2+y^2+4y+13-6x$
$=(x^2-6x+9)+(y^2+4y+4)$
$=(x-3)^2+(y+2)^2$
b.
$4x^2-4xy+1+2y^2-2y$
$=(4x^2-4xy+y^2)+(y^2-2y+1)$
$=(2x-y)^2+(y-1)^2$
c.
$x^2-2xy+2y^2+2y+1$
$=(x^2-2xy+y^2)+(y^2+2y+1)$
$=(x-y)^2+(y+1)^2$
a. \(x^2+y^2+4y+12-6x=\left(x^2-6x+9\right)+\left(y^2+4y+4\right)=\left(x-3\right)^2+\left(y+2\right)^2\)b. \(4x^2-4xy+1+2y^2-2y=\left(4x^2-4xy+y^2\right)+\left(y^2-2y+1\right)=\left(2x-y\right)^2+\left(y-1\right)^2\)c. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
a) x2+10x+26+y2+2y
=x2+10x+25+y2+2y+1
=(x+5)2+(y+1)2
b) z2-6z+5-t2-4t
=z2-6z+9-t2-4t-4
=(z-3)2-(t2+4t+4)
=(z-3)2-(t+2)2
c)x2-2xy+2y2+2y+1
=x2-2xy+y2+y2+2y+1
=(x-y)2+(y+1)2
d) 4x2-12x-y2+2y+8
=4x2-12x+9-y2+2y-1
=(2x-3)2-(y2-2y+1)
=(2x-3)2-(y-1)2
a)\(x^2+10x+26+y^2+2y\)
\(=x^2+10x+25+y^2+2y+1\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b)\(x^2-2xy+2y^2+1\)
\(=x^2-2xy+y^2+y^2+1\)
\(=\left(x-y\right)^2+y^2+1\)
c có lẽ sai ?
b:=y^2+2y+1+9x^2-12x+4
=(y+1)^2+(3x-2)^2
a:
SỬa đề: 5y^2
=y^2-10y+25+9x^2+4y^2-12xy
=(y-5)^2+(3x-2y)^2
1) x2 + 10x + 26 + y2 + 2y
= (x2 + 10x + 25) + (y2 + 2y + 1)
= (x2 + 5x + 5x + 25) + (y2 + y + y + 1)
= x(x + 5) + 5(x + 5) + y(y + 1) + (y + 1)
= (x + 5)2 + (y + 1)2
2) z2 - 6z + 13 + t2 + 4t
= (z2 - 6z + 9) + (t2 + 4t + 4)
= (z2 - 3z - 3z + 9) + (t2 + 2t + 2t + 4)
= z(z - 3) - 3(z - 3) + t(t + 2) + 2(t + 2)
= (z - 3)2 + (t + 2)2
3) x2 - 2xy + 2y2 + 2y + 1
(x2 - 2xy + y2) + (y2 + 2y + 1)
= (x - xy - xy + y2) + (y2 + y + y +1)
= x(x - y) - y(x - y) + y(y + 1) + (y + 1)
= (x - y)2 + (y + 1)2
a) \(x^2-4x+5+y^2+2y=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
b) \(2x^2+y^2-2xy+10x+25=\left(x^2+10x+25\right)+\left(x^2-2xy+y^2\right)\)
\(=\left(x+5\right)^2+\left(x-y\right)^2\)
c) \(2x^2+2y^2=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)=\left(x-y\right)^2+\left(x+y\right)^2\)
1)a)x2+10x+26+y2+2y
=(x2+10x+25)+(y2+2y+1)
=(x+5)2+(y+1)2
b)x2-2xy+2y2+2y+1
=(x2-2xy+y2)+(y2+2y+1)
=(x-y)2+(y+1)2
c)z2-6z+13+t2+4t
=(z2-6z+9)+(t2+4t+4)
=(z-3)2+(t+2)2
d)4x2+2z2-4xz-2z+1
=(4x2-4xz+z2)+(z2-2z+1)
=(2x-z)2+(z-1)2
2)a)(x-3)2-4=0
<=>(x-3-2)(x-3+2)=0
<=>(x-5)(x-1)=0
<=>x-5=0 hoặc x-1=0
<=>x=5 hoặc x=1
b)x2-2x=24
<=>x2-2x-24=0
<=>(x2-6x)+(4x-24)=0
<=>x(x-6)+4(x-6)=0
<=>(x-6)(x+4)=0
<=>x-6=0 hoặc x+4=0
<=>x=6 hoặc x=-4
a) x^2 + 10x + 26 + y^2 + 2y
=x2+10x+25+y2+2y+1
=x2+2.x.5+52+y2+2.y.1+12
=(x+5)2+(y+1)2
b)x^2 - 2xy + 2y^2 + 2y +1
=x2-2xy+y2+y2+2y+1
=(x-y)2+(y+1)2
c)z^2 - 6z + 13 + t^2 + 4t
=z2-6z+9+t2+4z+4
=z2-2.z.3+32+t2+2.t.2+22
=(z-3)2+(t+2)2
d)4x^2 + 2z^2 - 4xz - 2z + 1
=4x2-4xz+z2+z2-2z+1
=(2x)2-2.2x.z+z2+z2-2z.1+12
=(2x-z)2+(z-1)2
\(x^2+10x+26+y^2+2y=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=\left(x+5\right)^2+\left(y+1\right)^2\)
\(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
\(x^2+10x+26+y^2+2y=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=\left(x+5\right)^2+\left(y+1\right)^2\)