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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
\(\left(2x+y\right)^2+\left(2x-y\right)^2-5\left(x+y\right)\left(x-y\right)\)
\(=4x^2+4xy+y^2+4x^2-4xy+y^2-5x^2+5y^2\)
\(=3x^2+7y^2\)
Dễ mà bạn
\(P=\left(x-y\right)\left(x^2+xy+y^2\right)-2y^3=x^3-y^3-2y^3=x^3-3y^3=\left(\frac{1}{2}\right)^3-3.\left(\frac{2}{3}\right)^3=\frac{-55}{72}\)
a) A = (x + 2y)(x^2 - 2xy + 4y^2) - 8(x^3 + y^3)
A = x(x^2 - 2xy + 4y^2) + 2y(x^2 - 2xy + 4y^2) - 8(x^3 + y^3)
A = x^3 - 2x^2y + 4xy^2 + 2x^2y - 4xy^2 + 8y^3 - 8x^3 - 8y^3
A = -7x^3
b) B = (2x + y)^3 - (8x^3 + y^3) - 2x^2y
B = (2x + y)[(2x)^2 + 2.2xy + y^2] - 8x^3 - y^3 - 2x^2y
B = 2x[(2x)^2 + 2.2xy + y^2] + y[(2x)^2 + 2.2xy + y^3] - 8x^3 - y^3 - 2x^2y
B = 8x^3 + 8x^2y + 2xy^2 + 4x^2y + y^3 - 8x^3 - y^3 - 2x^2y
B = 10x^2y + 6xy^2
A= \(x^2+y^2+1+2x+2yx+2y=\left(x+y+1\right)^2\)
Chuc ban hoc tot
A=[(x-y).(x2+xy+y2)] +2y3
= x3-y3+2y3=x3+y3
=(2/3)3+(1/3)3
=4/9 + 1/9 =5/9
a/ \(\left(x-2y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(x-2y+3x-6y\right)=\left(x-2y\right)\left(4x-8y\right)\)
\(=4\left(x-2y\right)\left(x-2y\right)=4\left(x-2y\right)^2\)
b/ \(\left(y^2+1\right)\left(y+2\right)-\left(y+2\right)\left(y^2-2y+4\right)\)
\(=y^3+2y^2+y+2-y^3-8\)
\(=2y^2+y-6=2y^2+4y-3y-6\)
\(=\left(y+2\right)\left(2y-3\right)\)
riêng câu b mình có sửa đề lại, bn xem có đúng hong nha. Chúc bn hc tốt nhé ^^