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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
Ta có P + Q=x2 y + xy2 - 5x2 y2 + x3 + 3xy2 - x2 y + x2 y2
= -4x2 y2 + x3 + 4xy2
Chọn B
P + Q = (x2y + x3 – xy2 + 3) + (x3 + xy2 – xy – 6)
= x2y + x3 – xy2 + 3 + x3 + xy2 – xy – 6
= (x3 + x3) + x2y + (xy2 – xy2) – xy + (3 – 6)
= 2x3 + x2y – xy – 3
Vậy P + Q = 2x3 + x2y – xy – 3.
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/ \(P+Q=\left(x^2y+x^3-xy^2+3\right)+\left(x^3+xy^2-xy-6\right)\)
\(=x^2y+x^3-xy^2+3+x^3+xy^2-xy-6\)
\(=\left(x^3+x^3\right)+\left(xy^2-xy^2\right)+\left(3-6\right)+x^2y-xy\)
\(=2x^3+x^2y-xy-3\)
b/ \(M+N=\left(x^2y+0,5xy^3-7,5x^3y^2+x^3\right)+\)
\(\left(3xy^3-x^2y+5,5x^3y^2\right)\)
\(=x^2y+0,5xy^3-7,5x^3y^2+x^3+3xy^3-x^2y+5,5x^3y^2\)
\(=\left(x^2y-x^2y\right)+\left(0,5xy^3+3xy^3\right)+\left(5,5x^3y^2-7,5x^3y^2\right)+x^3\)
\(=3,5xy^3-2x^3y^2+x^3\)
a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)
\(=\left(x^2+y^2\right)\left(3x^2+2y^2\right)+y^2\)
\(=3x^2+3y^2=3\)
b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)
c: \(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(y-x\right)+3=3\)
d: \(=\left(x+y\right)^2-4\left(x+y\right)+1\)
=9-12+1
=-2
b)
\(A+B=\left(x^2y+2xy^2-7x^2y^2+x^4\right)+\left(5x^2y^2-2xy^2-x^2y-3x^4-1\right)\)
\(A+B=x^2y+2xy^2-7x^2y^2+x^4+5x^2y^2-2xy^2-x^2y-3x^4-1\)
\(A+B=(x^2y-x^2y)+(2xy^2-2xy^2)+(-7x^2y^2+5x^2y^2)+(x^4-3x^4)-1\)
\(A+B=-2x^2y^2-2x^4-1\)
c) \(-2.1^2.1^2-2.1^4-1=-3\)
CÂU C BẠN TÌM CÁCH LÀM NHA MIK KHÔNG BIẾT CÁCH TRÌNH BÀY
Ta có A + 2B = (x2y - xy2 + 3x2) + 2(x2y + xy2 - 2x2 - 1)
= x2y - xy2 + 3x2 + 2x2y + 2xy2 - 4x2 - 2
= 3x2y + xy2 - x2 - 2. Chọn C
a) Ta có: P = x2y + xy2 – 5x2y2 + x3 và Q = 3xy2 – x2y + x2y2
=> P + Q = x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 – 5x2y2 + x2y2 + x2y – x2y + xy2 + 3xy2
= x3 – 4x2y2 + 4xy2
b) Ta có: M = x3 + xy + y2 – x2y2 – 2 và N = x2y2 + 5 – y2.
=> M + N = x3 + xy + y2 – x2y2 – 2 + x2y2 + 5 – y2
= x3 – x2y2 + x2y2 + y2 – y2 + xy - 2 + 5
= x3 + xy + 3.