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a: A+B=x+2y+x-2y=2x
A-B=x+2y-x+2y=4y
b: A+B
=2x^2y-x^3-xy^2+1+x^3+xy^2-2
=2x^2y-1
A-B
=2x^2y-x^3-xy^2+1-x^3-xy^2+2
=-2x^3+2x^2y-2xy^2+3
c: A+B
=x^2-2yz+z^2+3yz+5x^2-z^2
=6x^2+yz
A-B
=x^2-2yz+z^2-3yz-5x^2+z^2
=-4x^2+2z^2-5yz
\(P=x^3+x^2y-5x^2-x^2y-xy^2+5xy+3\left(x+y\right)+2000\\ =x^2\left(x+y-5\right)-xy\left(x+y-5\right)+3\left(x+y-5\right)+2015\\ =x^2\left(5-5\right)-xy\left(5-5\right)+3\left(5-5\right)+2015\\ =2015\)
`P = x^3 + x^2 - 5x^2 - x^2y + xy^2 + 5xy + 3(x+y) + 2000`
`P = x^2(x+y) - (x+y)x^2 - xy(x+y) + (x+y)xy + 3(x+y) + 2000`
`P = 0 + 0 + 3.5 + 2000`
`P = 2015`
\(P=\dfrac{1}{3}x^2y+xy^2-xy+\dfrac{1}{2}xy^2-5xy-\dfrac{1}{3}x^2y=\dfrac{3}{2}xy^2-6xy\)
Thay x = 2 ; y = 1 ta được
\(\dfrac{3}{2}.2.1-6.2.1=3-12=-9\)
Bài 2: Tính giá trị của biểu thức:
a) P= 1/3 x^2 y + xy^2 - xy + 1/2 xy^2 - 5xy - 1/3 x^2 y (1)
Tại x = 0,5; y = 1
Thay \(x=0,5 ; y=1\) vào biểu thức (1) , ta có :
P= \(\dfrac{1}{3} . 0,5^2.1+0,5.1^2-0,5.1+\dfrac{1}{2}. 0,5.1^2-5.0,5.1-\dfrac{1}{3}.0,5^2.1\)
P= \(=\dfrac{1}{12}+\dfrac{1}{2} -0,5+\dfrac{1}{4} -\dfrac{5}{2} - \dfrac{1}{12}\)
P= \(= \dfrac{-9}{4}\)
Vậy \(P =\dfrac{-9}{4}\)
\(Q=5x^2y-3xy+\dfrac{1}{2}x^2y-xy+5xy-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}x-\dfrac{1}{4}\)
\(Q=\left(5-3+\dfrac{1}{2}+5-\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{1}{4}\right)+\left(x^2xx^2xxx\right)+\left(yyyyy\right)+\left(-x\right)\)
\(Q=\dfrac{97}{12}+x^{^{ }8}+y^5+\left(-x\right)\)
\(Q=\dfrac{97}{12}+x^7+y^5\)
\(Q=x^2+2xy+\left(-3x^3+3x^3\right)+\left(2y^3-y^3\right)=x^2+2xy+y^3\)
\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)
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) \(\left(x^2+y^2+z^2\right)+\left(x^2-y^2+z^2\right)\)
= \(x^2+y^2+z^2+x^2-y^2+z^2\)
=\(2x^2+2z^2\)
=\(2\left(x+z\right)^2\)