Bài 27: Tìm đa tức M, biết:
a. M + ( 5×2 – 2xy ) = 6×2+ 9xy – y2
b. M – (3xy – 4y2) = x2 -7xy + 8y2
c. (25x2y – 13 xy2 + y3) – M = 11x2y – 2y2;
d. M + ( 12×4 – 15x2y + 2xy2 +7 ) = 0
Bài 27: Tìm đa tức M, biết:
a. M + ( 5×2 – 2xy ) = 6×2+ 9xy – y2
b. M – (3xy – 4y2) = x2 -7xy + 8y2
c. (25x2y – 13 xy2 + y3) – M = 11x2y – 2y2;
d. M + ( 12×4 – 15x2y + 2xy2 +7 ) = 0
Đáp án:
a , Ta có :
M + ( 5×2 – 2xy ) = 6×2+ 9xy – y2
=> M = 6×2+ 9xy – y2 – ( 5×2 – 2xy )
=> M =6×2+ 9xy – y2 – 5×2 + 2xy
=> M = ( 6×2 – 5×2 ) + ( 9xy + 2xy ) – y2
=> M = x2 + 11xy – y2
b, M – (3xy – 4y2) = x2 -7xy + 8y2
=> M = x2 -7xy + 8y2 + 3xy – 4y2
=> M = ( 8y2 – 4y2) – (7xy – 3xy ) + x2
=> M = 4y2 – 4xy + x2
c, (25x2y – 13 xy2 + y3) – M = 11x2y – 2y2
=> M = 25x2y – 13 xy2 + y3 – (11x2y – 2y2)
=> M = 25x2y – 13 xy2 + y3 – 11x2y + 2y2
=> M = ( 25x2y – 11x2y ) + y3 – 13xy2 + 2y2
=> M = 14x2y + y3 – 13xy2 + 2y2
d,. M + ( 12×4 – 15x2y + 2xy2 +7 ) = 0
=> M = – ( 12×4 – 15x2y + 2xy2 +7 )
=> M = -12×4 + 15x2y – 2xy2 – 7
Giải thích các bước giải:
$a$) $M + (5x^2 – 2xy) = 6x^2 + 9xy – y^2$
$⇔ M = 6x^2 + 9xy – y^2 – 5x^2 + 2xy$
$⇔ M = x^2 + 11xy – y^2$
$b$) $M – (3xy – 4y^2) = x^2 – 7xy + 8y^2$
$⇔ M = x^2 – 7xy + 8y^2 + 3xy – 4y^2$
$⇔ M = x^2 – 4xy + 4y^2$
$c$) $(25x^2y – 13xy^2 + y^3) – M = 11x^2 y – 2y^2$
$⇔ M = 25x^2y – 13xy^2 + y^3 – 11x^2y + 2y^2$
$⇔ M = 14x^2y – 13xy^2 + y^3 + 2y^2$
$d$) $M + ( 12x^4 – 15x^2y + 2xy^2 +7 ) = 0$
$⇔ M = -12x^4 + 15x^2y – 2xy^2 – 7$