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\(\text{Ta có: }\)
\(=-5xy^2+7xy^2+(-15xy^2)\)
\(=xy^2.(-5+7-15)\)
\(=-13xy^2\)
\(.a.\)
\(2x^2+3x^2-7x^2\)
\(=\left(2+3-7\right).x^2\)
\(=\left(-2\right).x^2\)
\(.b.\)
\(5xy-\frac{1}{3}xy+xy\)
\(=\left(5-\frac{1}{3}+1\right).xy\)
\(=\frac{17}{3}.xy\)
\(.c.\)
\(15xy^2-\left(-5xy^2\right)\)
\(=15xy^2+5xy^2\)
\(=\left(15+5\right).xy^2\)
\(=20xy^2\)
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
a, = 3x^2y^2.4x^2y^2 = 12x^4y^4
b, = xy^3.(4+5-6) = 3.xy^3
Tk mk nha
a) Ta có M = x2y + 0,5xy3 – 7,5x3y2 + x3 và N = 3xy3 – x2y + 5,5x3y2.
=> M + N = x2y + 0,5xy3 – 7,5x3y2 + x3 + 3xy3 – x2y + 5,5x3y2
= – 7,5x3y2 + 5,5x3y2 + x2y – x2y + 0,5xy3 + 3xy3 + x3
= -2x3y2 + 3,5xy3 + x3
b) P = x5 + xy + 0,3y2 – x2y3 – 2 và Q = x2y3 + 5 – 1,3y2.
=> P + q = (x5 + xy + 0,3y2 – x2y3 – 2) + (x2y3 + 5 – 1,3y2)
= x5 + xy + 0,3y2 – x2y3 – 2 + x2y3 + 5 – 1,3y2
= x5 – x2y3 + x2y3 + 0,3y2 – 1,3y2 + xy - 2 + 5
= x5 - y2 + xy + 3.
a) (5x2y-5xy2+xy) + (xy-x2y2+5xy2)
= 5x2y-5xy2+xy+xy-x2y2+5xy2
= 5x2y+(5xy2-5xy2)+(xy+xy)-x2y2
= 5x2y+2xy-x2y2
b) (x2+y2+z2) + (x2-y2+z2)
= x2+y2+z2+x2-y2+z2
= (x2+x2)+(y2-y2)+(z2+z2)
= 2x2+2z2
a)( \(5x^2y\)\(-\) \(5xy^2\) \(+\) \(xy\)) + (\(xy\) \(-\) \(x^2y^2\) \(+\) \(5xy^2\))
= \(5x^2y-5xy^2+xy+xy-x^2y^2+5xy^2\)
= \(5x^2y+2xy-x^2y^2\)
b) \(\left(x^2+y^2+z^2\right)+\left(x^2-y^2+z^2\right)\)
= \(x^2+y^2+z^2+x^2-y^2+z^2\)
=\(2x^2+2z^2\)
=\(2\left(x+z\right)^2\)
Ta có: xy3 + 5xy3 + (-7xy3) = (3 + 5 - 7) xy3 = 1. xy3 = xy3