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\(B=x^5y^2+\dfrac{1}{2}x^5y^2-6xy+1=\dfrac{3}{2}x^5y^2-6xy+1\)
a) 2x2yz + 4xy2z - 5x2yz + xy2z - xyz
= (2x2yz-5x2yz)+(4xy2z+xy2z)-xyz
= -3x2yz + 5xy2z - xyz
b) x3-5xy+3x3+xy-x2+\(\dfrac{1}{2}\)xy-x2
= (x3+3x3)+(xy-5xy+\(\dfrac{1}{2}\)xy)-(x2+x2)
= 4x3-\(\dfrac{7}{2}\)xy-2x2
a: \(A=\left(5xy-2xy+4xy\right)+3x-2y-y^2\)
\(=7xy+3x-2y-y^2\)
b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2-\dfrac{1}{2}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)
\(=\dfrac{-7}{8}ab^2+\dfrac{3}{8}a^2b\)
c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)
\(=7a^2b-11b^2+9c^2\)
\(A=5xy-y^2-2xy+4xy+3x-2y\)
\(A=-y^2+7xy+3x-2y\)
\(B=\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2+\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b-\dfrac{1}{2}ab^2\)
\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)
\(C=2a^2b-8b^2+5a^2b+5c^2-3b^2+4c^2\)
\(C=7a^2b-11b^2+9c^2\)
Thu gọn rồi tính giá trị của đa thức P tại x = 0,5 và y = 1.
Ta có: P = 1313 x2 y + xy2 – xy + 1212 xy2 – 5xy – 1313 x2y
P = 1313 x2 y – 1313 x2y + 1212 xy2 + xy2 – xy – 5xy = 3232 xy2 – 6xy
Thay x = 0,5 và y = 1 ta được
P = 3232 . 0,5 . 12 – 6. 0,5 . 1 = 3434 - 3 = −94−94.
Vậy P = −94−94 tại x = 0,5 và y = 1.
a: \(=-2\cdot\dfrac{1}{4}\cdot3\cdot x^2y^2\cdot xy\cdot xy^3=-\dfrac{3}{2}x^4y^6\)
b: \(=4x^6y^2\cdot xy^2\cdot\dfrac{1}{2}y^5=2x^7y^9\)
a) \(=\left(2.\dfrac{1}{4}.\left(-3\right)\right).\left(x^2.x.x\right).\left(y^2.y^3.y\right)\)
\(=-\dfrac{3}{2}x^4y^6\)
b) \(-4x^6y^2.x.y^2.\dfrac{1}{2}y^5\)
\(=\left(4.\dfrac{1}{2}\right).\left(x^6.x\right).\left(y^2.y^2.y^5\right)\)
\(=2x^7y^9\)
1.
a)\(\left(\dfrac{1}{2}\cdot\left(-2\right)\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)
\(\dfrac{1}{3}x^6y^5z\)
Deg=12
Thu gọn đa thức:
\(C=-\dfrac{1}{2}x^2y-2xy+\dfrac{1}{2}x^2y-xy+xy-\dfrac{1}{3}x+\dfrac{1}{2}+x-0,25\)
\(=x^2y\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+xy\left(-2-1+1\right)+x\left(-\dfrac{1}{3}+1\right)+\dfrac{1}{2}-\dfrac{1}{4}\)
\(=-2xy+\dfrac{2}{3}x+\dfrac{1}{4}\)
\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)
Bậc:3
Thay x=-1, y=1 vào B ta có:
\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)
a: \(A=\dfrac{3}{2}x^2+6x+3\)
b: \(B=5xy-\dfrac{2}{3}xy+\dfrac{1}{2}x^2y+2x^2y=\dfrac{5}{2}x^2y+\dfrac{13}{3}xy\)
a) \(2x^2+x-\dfrac{1}{2}x^2+5x+3\)\(\)
= \(\left(2x-\dfrac{1}{2}x^2\right)+\left(x+5x\right)+3\)
= \(\dfrac{3}{2}x^2+6x+3\)
Vậy A = \(\dfrac{3}{2}x^2+6x+3\)