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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. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
\(a.\)
\(z^2-6z+5-t^2-4t\)
\(=z^2-6z+9-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
\(b.\)
\(4x^2-12x-y^2+2y+1\)
Câu này đề sai sao ấy em !
b, mik nghĩ đề sửa thành: \(4x^2-12x-y^2+2y+8\)
\(=4x^2-12x+9-y^2+2y-1\)
\(=\left(2x\right)^2-2.2.3.x+3^2-\left(y^2-2y+1\right)\)
\(=\left(2x-3\right)^2-\left(y-1\right)^2\)
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
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)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
x2 – 2xy + y2 – z2 + 2zt – t2
(Nhận thấy x2 – 2xy + y2 và z2 – 2zt + t2 là các hằng đẳng thức)
= (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2 (xuất hiện hằng đẳng thức (3))
= [(x – y) – (z – t)][(x – y) + (z – t)]
= (x – y – z + t)(x – y + z –t)
1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2
b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2
c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2
2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16
= x2 + 2xy + y2 + 42 = (x + y)2 + 42
b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36
= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2
c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9
= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2
d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2
= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2
a) \(x^2+2x+1=x^2+2\cdot x\cdot1+1^2=\left(x+1\right)^2\)
b) \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)
c) \(x^2+6xy+9y^2=x^2+2\cdot x\cdot3y+\left(3y\right)^2=\left(x+3y\right)^2\)
d) \(z^2-z+\dfrac{1}{4}=z^2-2\cdot z\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(z-\dfrac{1}{2}\right)^2\)
e) \(25x^2-10x+1=\left(5x\right)^2-2\cdot5x\cdot1+1^2=\left(5x-1\right)^2\)
\(1.z^2-6z+5-t^2-4t\)
\(=\left(z^2-6z+9\right)-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
\(3,x^2-2xy+2y^2+2y+1\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
\(x^2+2y^2-2xy+2y+1\)
\(=x^2-2xy+y^2+y^2+2y+1\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
câu b vs c nữa