Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x\left(xy+1\right)+y\left(xy-1\right)-xy\left(x+y\right)\)
\(=X^2y+x+xy^2-y-x^2y-xy^2\)
\(=x-y\)
x+y+1=0 suy ra x+y=1
Làm câu A nhé B,C tương tự
A= x^2.(x+y-2)-(xy+y^2-2y)+(y+x-1)=0-y.(x+y-2)+1=1
Hok tốt
A = 5x(x - y) - y(5x - y)
A = 5x2 - 5xy - 5xy + y2
A = 5x2 - 10xy + y2 (1)
Thay x = -1; y = 3 vào (1), ta có:
5.(-1)2 - 10.(-1).3 + 32 = 44
B = 4y(x2 - 3xy + 3y2) - 2xy(2x - 6y - 3)
B = 4x2y - 12x2 + 12y3 - 4x2y + 12xy2 + 6xy
B = 12y3 + 6xy (1)
Thay x = 5; y = -1 vào (1), ta có:
12.(-1)3 + 6.5.(-1) = -42
C = 5x2(x - y2) + 3x(xy2 - y) - 5x3
C = 5x3 - 5x2y2 + 3x2y2 - 3xy - 5x3
C = -2x2y2 - 3xy (1)
Thay x = -2; y = -5 vào (1), ta có:
-2.(-2)2.(-5)2 - 3.(-2).(-5) = -230
D = 6x2(y2 - xy + 2x2y) - 3xy(2xy - x2 + 4x3)
D = 6x2y2 - 6x3y + 12x4y - 6x2y2 + 3x3y - 12x4y
D = -3x3y (1)
Thay x = 11; y = -1 vào (1), ta có:
-3.113.(-1) = 3993
a: \(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{8}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)
\(=\dfrac{-1}{8}+\dfrac{1}{12}-\dfrac{1}{18}=-\dfrac{7}{72}\)
b: \(B=\left(-1\cdot3\right)^2+\left(-1\right)\cdot3-1+27\)
\(=9-3-1+27\)
=36-4=32
c: \(C=-0.7xy^2-2x^2y-4.5xy\)
\(=-0.7\cdot\dfrac{1}{2}\cdot1-2\cdot0.25\cdot\left(-1\right)-4.5\cdot0.5\cdot\left(-1\right)\)
\(=\dfrac{-7}{20}+\dfrac{1}{2}+\dfrac{9}{2}\cdot\dfrac{1}{2}\)
\(=\dfrac{12}{5}\)
a) \(\left(-\frac{1}{2}x^3y\right)^2\cdot2xy\cdot\left(-xy\right)^2=\left(-\frac{1}{2}\right)^2x^6y^2\cdot2xy\cdot\left(-1\right)^2x^2y^2\)
\(=\frac{1}{4}x^6y^2\cdot2xy\cdot x^2y^2=\left(\frac{1}{4}\cdot2\right)x^6x\cdot x^2\cdot y^2\cdot y\cdot y^2=\frac{1}{2}x^9y^5\)
b) \(\left(\frac{1}{3}x^3y\right)\left(xy^2\right)^2\cdot\frac{3}{2}x^2=\frac{1}{3}x^3y\cdot x^2y^4\cdot\frac{3}{2}x^2\)
\(=\left(\frac{1}{3}\cdot\frac{3}{2}\right)x^3\cdot x^2\cdot x^2\cdot y\cdot y^4=\frac{1}{2}x^7y^5\)
\(=\frac{1}{4}x^6y^2\cdot2xy\cdot x^2y^2=\left(\frac{1}{4}\cdot2\right)x^6x\cdot x^2\cdot y^2\cdot y\cdot y^2=\frac{1}{2}x^9y^5\)