Cho hai đa thức P = x2y2 - 4x2y - xy2 + 2xy và Q = 4x2y2 + xy; Tính P + Q = ?
A) 5x2y2 - 4x2y - xy2 + 3xy
B) x2y2 + 3xy
C) 5x2y2 - 4x2y - xy2 + xy
D) x2y2 - 4x2y - xy2 + 3xy
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d) \(4x^2y^2-8xy^2+4y^2=4y^2\left(x^2-2x+1\right)=4y^2\left(x-1\right)^2\)
e) \(x^3y+10x^2y+35xy=xy\left(x^2+10x+35\right)\)
f) \(2x^3-4x^2y+2xy^2-8x=2x\left(x^2-2xy+y^2-4\right)=2x\left[\left(x-y\right)^2-4\right]=2x\left(x-y-2\right)\left(x-y+2\right)\)
g) \(3x^2-9xy-6x+18y=3x\left(x-3y\right)-6\left(x-3y\right)=3\left(x-3y\right)\left(x-2\right)\)
h) \(x^2y^2-3xy^2+2xy-6y=xy\left(xy+2\right)-3y\left(xy+2\right)=\left(xy+2\right)\left(xy-3y\right)=y\left(xy+2\right)\left(x-3\right)\)
d: \(4x^2y^2-8xy^2+4y^2\)
\(=4y^2\left(x^2-2x+1\right)\)
\(=4y^2\left(x-1\right)^2\)
e: \(x^3y+10x^2y+35xy\)
\(=xy\left(x^2+10x+35\right)\)
f: \(2x^3-4x^2y+2xy^2-8x\)
\(=2x\left(x^2-2xy+y^2-4\right)\)
\(=2x\left(x-y-2\right)\left(x-y+2\right)\)
a: \(4x^3-xy^2\)
\(=x\left(4x^2-y^2\right)\)
\(=x\left(2x-y\right)\left(2x+y\right)\)
b: \(5x^3-10x^2+5x\)
\(=5x\left(x^2-2x+1\right)\)
\(=5x\left(x-1\right)^2\)
c: \(4x^2+24x+36-4y^2\)
\(=4\left(x^2+6x+9-y^2\right)\)
\(=4\left(x+3-y\right)\left(x+3+y\right)\)
Ta có P + Q=x2 y + xy2 - 5x2 y2 + x3 + 3xy2 - x2 y + x2 y2
= -4x2 y2 + x3 + 4xy2
Chọn B
Ta có:
C(x) = (5x2y - 4xy2 + 5x - 3) - (xyz - 4x2y + xy2 + 5x - 1)
= 5x2y - 4xy2 + 5x - 3 - xyz + 4x2y - xy2 - 5x + 1
= -xyz + 9x2y - 5xy2 - 2
Chọn C
a: Bậc của M là 4
Bậc của N là 4
b: N+K=M nên K=M-N
\(=x^2y^2-4x^2y-4xy^2+6xy+10-x^2y^2-6xy-10\)
\(=-4x^2y-4xy^2\)
a, 2xy +2x2 - 4xy2 - 2 ; b, -3x2y2 -2x2y + y ; c, 3x3 - 2y - 3
Ta có: P = x2y + xy2 – 5x2y2 + x3 và Q = 3xy2 – x2y + x2y2
⇒ P + Q = (x2y + xy2 – 5x2y2 + x3) + (3xy2 – x2y + x2y2)
= x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 +(– 5x2y2 + x2y2)+ (x2y – x2y) + (xy2+ 3xy2)
= x3 – 4x2y2 + 0 + 4xy2
= x3 – 4x2y2 + 4xy2
11: \(\dfrac{1}{3}x^2y^2\left(6x+\dfrac{2}{3}x^2-y\right)\)
\(=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\)
12: \(\dfrac{3}{4}x^3y^2\left(4x^2y-x+y^5\right)\)
\(=3x^5y^3-\dfrac{3}{4}x^4y^2+\dfrac{3}{4}x^3y^7\)
13: \(-5x^2y^4\left(3x^2y^3-2x^3y^2-xy\right)\)
\(=-15x^4y^7+10x^5y^6+5x^3y^5\)
Ta có:
• P + Q = (x2y + x3 – xy2 + 3) + (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 + x3 + xy2 – xy – 6
= x2y + (x3 + x3) + (xy2 – xy2) – xy + (3 – 6)
= x2y + 2x3 – xy – 3.
• P – Q = (x2y + x3 – xy2 + 3) – (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 – x3 – xy2 + xy + 6
= x2y + (x3 – x3) – (xy2 + xy2) + xy + (6 + 3)
= x2y – 2xy2 + xy + 9.
Vậy P + Q = x2y + 2x3 – xy – 3; P – Q = x2y – 2xy2 + xy + 9.
\(\text{ P + Q = (x^2y + x^3 – xy^2 + 3) + (x^3 + xy^2 – xy – 6)}\)
\(\text{= x^2y + x^3 – xy^2 + 3 + x^3 + xy^2 – xy – 6}\)
\(\text{= x^2y + (x^3 + x^3) + (xy^2 – xy^2) – xy + (3 – 6)}\)
\(\text{= x^2y + 2x^3 – xy – 3}\)
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\(\text{P – Q = (x^2y + x^3 – xy^2 + 3) – (x^3 + xy^2 – xy – 6)}\)
\(\text{= x^2y + x^3 – xy^2 + 3 – x^3 – xy^2 + xy + 6}\)
\(\text{= x^2y + (x^3 – x^3) – (xy^2 + xy^2) + xy + (6 + 3)}\)
\(\text{= x^2y – 2xy^2 + xy + 9}\)
A
A