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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. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
Bài 1 :
a ) \(x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right)\left(x^3+2\right)\)
b ) \(-9x^2+1=1-\left(3x\right)^2=\left(1-3x\right)\left(1+3x\right)\)
c ) \(x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right)\left(x^5+3\right)\)
Bài 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 ) \(x^2-2xy+2y^2+2y+1\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c ) \(a^2-6a+10a+b^2+4b\)
\(=a^2+4a+b^2+4b\)
\(=a^2+4a+4+b^2+4b+4-8\)
\(=\left(a+2\right)^2+\left(b+2\right)^2-8\)
d ) \(4x^2+2y^2-4xy-2y+1\)
\(=\left(2x\right)^2-4xy+y^2+y^2-2y+1\)
\(=\left(2x-y\right)^2+\left(y-1\right)^2\)
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
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
a) \(x^4+4x^2+4=\left(x^2+2\right)^2\)
b) \(\left(2y-x\right)^2+2\left(2y-x\right)+1=\left(2y-x+1\right)^2\)
c) \(\left(2a-4b\right)^2+4a-8b+1=\left(2a-4b\right)^2+2\cdot\left(2a-4b\right)\cdot1+1^2=\left(2a-4b+1\right)^2\)
a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)
\(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+y^2+2xy=\left(x+y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)
\(a.\)
\(z^2-6z+5-t^2-4t\)
\(=z^2-6z+9-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
\(b.\)
\(4x^2-12x-y^2+2y+1\)
Câu này đề sai sao ấy em !
b, mik nghĩ đề sửa thành: \(4x^2-12x-y^2+2y+8\)
\(=4x^2-12x+9-y^2+2y-1\)
\(=\left(2x\right)^2-2.2.3.x+3^2-\left(y^2-2y+1\right)\)
\(=\left(2x-3\right)^2-\left(y-1\right)^2\)
`a)x^2-2x+2+4y^2+4y`
`=x^2-2x+1+4y^2+4y+1`
`=(x-1)^2+(2y+1)^2`
`b)4x^2+y^2+12x+4y+13`
`=4x^2+12x+9+y^2+4y+4`
`=(2x+3)^2+(y+2)^2`
`c)x^2+17+4y^2+8x+4y`
`=x^2+8x+16+4y^2+4y+1`
`=(x+4)^2+(2y+1)^2`
`d)4x^2-12xy+y^2-4y+13`
`=4x^2-12x+9+y^2-4y+4`
`=(2x-3)^2+(y-2)^2`
a) \(x^2-2x+2+4y^2+4y=\left(x-1\right)^2+\left(2y+1\right)^2\)
b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)
c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)
d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)
a) \(x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right).\left(x^3+2\right)\)
b) \(-9x^2+1=1^2-\left(3x\right)^2=\left(1-3x\right).\left(1+3x\right)\)
c) \(x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right).\left(x^5+3\right)\)
mk chỉ làm đk bài 1 thui ,thông cảm cho mk nha bạn
\(a;x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right)\left(x^3+2\right)\)
\(b;-9x^2+1=1^2-3x^2=\left(1-3x\right).\left(1+3x\right)\)
\(c;x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right).\left(x^5+3\right)\)
\(#LTH\)