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6 tháng 9 2023
m: (x-y)(x^2-2xy+y^2)
=(x-y)*(x-y)^2
=(x-y)^3
=x^3-3x^2y+3xy^2-y^3
n: =-(x^3+x^2y-x-x^2y-xy^2+y)
=-x^3+x+xy^2-y
o: =-(x^3+x^2y^2-x^2-2xy-2y^3+2y)
=-x^3-x^2y^2+x^2+2xy+2y^3-2y
p: (1/2x-1)(2x-3)
=1/2x*2x-1/2x*3-2x+3
=x^2-3/2x-2x+3
=x^2-7/2x+3
q: (x-1/2y)(x-1/2y)
=(x-1/2y)^2
=x^2-xy+1/4y^2
r: (x^2-2x+3)(1/2x-5)
=1/2x^3-5x^2-x^2+10x+3/2x-15
=1/2x^3-6x^2+11,5x-15
S
16 tháng 8 2016
\(B=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)
\(=\frac{x^2y-x^2z+zy^2-xy^2+z^2x-z^2y}{x^2\left(y-z\right)-y^2\left(y-z\right)}\)
\(=\frac{\left(x^2y-z^2y\right)-\left(xy^2-zy^2\right)-\left(x^2z-z^2x\right)}{\left(x^2-y^2\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x+z\right)-y^2-xz\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(xy+zy-y^2-xz\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[\left(xy-y^2\right)-\left(xz-zy\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left[y\left(x-y\right)-z\left(x-y\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)
\(=\frac{x-z}{x+y}\)
S
16 tháng 8 2016
\(A=\frac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
\(=\frac{x^2y-y^2+x^2-y+x^2y^2-1}{x^2y+y^2+x^2+y+x^2y^2+1}\)
\(=\frac{\left(x^2y+x^2\right)+\left(x^2y^2-y^2\right)-\left(y+1\right)}{\left(x^2y+x^2\right)+\left(x^2y^2+y^2\right)+\left(y+1\right)}\)
\(=\frac{x^2\left(y+1\right)+y^2\left(x^2-1\right)-\left(y+1\right)}{x^2\left(y+1\right)+y^2\left(x^2+1\right)+\left(y+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y+1\right)+y^2\left(x^2-1\right)}{\left(x^2+1\right)\left(y+1\right)+y^2\left(x^2+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(y^2+y+1\right)}{\left(x^2+1\right)\left(y^2+y+1\right)}\)
\(=\frac{x^2-1}{x^2+1}\)
15 tháng 6 2018
+) \(A=x^2-y+xy^2-x\)
\(A=\left(x^2-y\right)+\left(xy^2-x\right)\)
\(A=\left(x^2-y\right)+x\left(y^2-1\right)\)
Tại x = -5, y = 2 ta có :
\(A=\left[\left(-5\right)^2-2\right]+\left(-5\right)\left(2^2-1\right)=8\)
+) \(B=3x^3-2y^3-6x^2y^2\)
\(B=3x^3-\left(2y^3+6x^2y^2\right)=3x^3-2y^2\left(y+3x^2\right)\)
Tại x = 2/3, y = 1/2 ta có :
\(B=3.\left(\dfrac{2}{3}\right)^3-2.\left(\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}+3.\dfrac{4}{9}\right)=\dfrac{55}{36}\)
+) \(C=2x+xy^2-x^2y-y\)
\(C=\left(2x+xy^2\right)-\left(x^2y+y\right)=x\left(2+y^2\right)-y\left(x^2+1\right)\)
Tại x= -1/2, y = -1/3 ta có :
\(C=\left(\dfrac{-1}{2}\right)\left[2+\left(\dfrac{-1}{3}\right)^2\right]-\left(-\dfrac{1}{3}\right)\left[\left(\dfrac{-1}{2}\right)^2+1\right]=\left(-\dfrac{19}{18}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{-23}{36}\)
11 tháng 9 2020
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=\left(x^2-4y^2\right)^2\)
\(=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
A
11 tháng 9 2020
\(\left(2x+y\right)^2=4x^2+4xy+y^2\)
\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)
\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{x^2y}{2}+\frac{x^2y}{2}-\frac{y^2}{4}=x^4-\frac{y^2}{4}\)
\(\left(x-2y\right)^2\left(x+2y\right)^2=x^4-8x^2y^2+16y^4\)
\(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-xy^2+xy^2-1=x^2y^4-1\)
\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)
