(x + 9y / x^2 - 9y^2 - 3y / x^2 + 3xy) . x - 3xy / x + 3y
Mn help me câu này vs:(x-3y)(x^2+3xy+9y^2)+(x+3y)(x^2-3xy+9y^2)tại x=-1 và y=-2019
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)+\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
\(=x^3-27y^3+x^3+27y^3=2x^3=2.\left(-1\right)^3=-2\)
\(\Leftrightarrow x^3-3y^3+x^3+3y^3\)
\(\Leftrightarrow2x^3\)
chứng minh giá trị của biểu thức sau không phụ thuộc vào giá trị của biến A=(x+3y) ( x^2 - 3xy +9y^2) + 3y(x+3y)(x-3y)-x(3xy+x^2 -5) -5x+1
\(A=(x+3y)(x^2-3xy+9y^2)+3y(x+3y)(x-3y)-x(3xy+x^2-5)-5x+1\\A=(x+3y)[x^2-x\cdot3y+(3y)^2]+3y[x^2-(3y)^2]-3x^2y-x^3+5x-5x+1\\A=x^3+(3y)^3+3y(x^2-9y^2)-3x^2y-x^3+1\\A=x^3+27y^3+3x^2y-27y^3-3x^2y-x^3+1\\A=1\)$\Rightarrow$ Giá trị của $A$ không phụ thuộc vào giá trị của biến.
c/m dang thuc : (x^2 +3xy)/(x^2 - 9y^2) + (2x^2 -5xy-3y^2)/(x^2-6xy+9y^2)= (3x^2 +2xy+3xz +6yz)/(xz -3yz +z^2-3xy)
\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
đk: \(x\ne0\); \(x\ne\pm3y\)
\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
\(=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
\(=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x+3y}{x\left(x-3y\right)}\)
c/m dang thuc : (x^2 +3xy)/(x^2 - 9y^2) + (2x^2 -5xy-3y^2)/(x^2-6xy+9y^2)= (3x^2 +2xy+3xz +6yz)/(xz -3yz +z^2-3xy)
a,(2x+3)^3 b,(x-3y)^3 c.(x+4)(x^2-4x+15) d,(1/3x+1y)(1/9x^2-2/3xy+4y) e,(x-3y)(x^2+3xy+9y^2)
a: \(\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
b: \(\left(x-3y\right)^3=x^3-9x^2y+27xy^2-27y^3\)
a) (x^3 - 2y)^3
b) (x-3y) (x^2 + 3xy +9y^2)
a) \(\left(x^3-2y\right)^3\)
\(=\left(x^3-2y\right)\left[\left(x^3\right)^2+x^3.2y+\left(2y^2\right)\right]\)
\(=\left(x^3-2y\right)\left(x^6+2x^3y+4y^2\right)\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left(x^2+x.3y+\left(3y\right)^2\right)\)
\(=\left(x-3y\right)^3\)
\(\dfrac{x+9y}{x^2-9y^2}\) - \(\dfrac{3y}{x^2+3xy}\)
\(\dfrac{x+9y}{x^2-9y^2}-\dfrac{3y}{x^2+3xy}\)
\(=\dfrac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\dfrac{3y}{x\left(x+3y\right)}\)
\(=\dfrac{x\left(x+9y\right)-3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\dfrac{x-3y}{x\left(x+3y\right)}\)
(2xy: x^2 - y^2 + x-y : 2x + 2y) : x+y:2x + y:y-x
x^2+3xy: x^2 - 9y^2 + 2x^2 - 5xy- 3y^2 : 6xy - x^2- 9y^2 - x^2+ xz + xy + yz: 3yz - x^2 - xz + 3xy