Rút gọn P=x2y+x3-xy2+3+x3+xy2-xy-6
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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/ \(M+N=\left(x^2y+0,5xy^3-7,5x^3y^2+x^3\right)+\)
\(\left(3xy^3-x^2y+5,5x^3y^2\right)\)
\(=x^2y+0,5xy^3-7,5x^3y^2+x^3+3xy^3-x^2y+5,5x^3y^2\)
\(=\left(x^2y-x^2y\right)+\left(0,5xy^3+3xy^3\right)+\left(5,5x^3y^2-7,5x^3y^2\right)+x^3\)
\(=3,5xy^3-2x^3y^2+x^3\)
\(A=5x^2y-xy^2+4xy+6\) bậc : 3
a)\(B=-5x^2y+xy^2-4xy-6\)
b)\(=>C=-2xy+1-5x^2y+xy^2-4xy-6\)
\(C=-5x^2y+xy^2-6xy-5\)
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
\(a,=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\\ b,=\left(2x^3-6x^2+10x-3x^2+9x-15\right):\left(x^2-3x+5\right)\\ =\left[2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)\right]:\left(x^2-3x+5\right)\\ =2x-3\\ c,=\left[x^2\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)=x^2+1\)
\(x^2y+xy^2+x+y=240\)
\(\Leftrightarrow xy\left(x+y\right)+x+y=240\)
\(\Leftrightarrow11\left(x+y\right)+x+y=240\)
\(\Rightarrow12\left(x+y\right)=240\)
\(\Rightarrow x+y=20\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=20^3-3.11.20=\)
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
Với x ≥ 0; y ≥ 0 thì x + y ≥ 0
Ta có: x3 + y3 ≥ x2y + xy2
⇔ (x3 + y3) – (x2y + xy2) ≥ 0
⇔ (x + y)(x2 – xy + y2) – xy(x + y) ≥ 0
⇔ (x + y)(x2 – xy + y2 – xy) ≥ 0
⇔ (x + y)(x2 – 2xy + y2) ≥ 0
⇔ (x + y)(x – y)2 ≥ 0 (Luôn đúng vì x + y ≥ 0 ; (x – y)2 ≥ 0)
Dấu « = » xảy ra khi (x – y)2 = 0 ⇔ x = y.
x2y+x3-xy2+3+xy2-xy-6=x2y+2x3-3