Có: (2x + y)(4x2 - 2xy + y2) = 8x3 + y3
Lời giải:
Ta có: (2x + y)(4x2 - 2xy + y2)
= 2x(4x2 - 2xy + y2) + y(4x2 - 2xy + y2)
= 8x3 - 4x2y + 2xy2 + 4x2y - 2xy2 + y3
= 8x3 + y3.
Có: (2x + y)(4x2 - 2xy + y2) = 8x3 + y3
Lời giải:
Ta có: (2x + y)(4x2 - 2xy + y2)
= 2x(4x2 - 2xy + y2) + y(4x2 - 2xy + y2)
= 8x3 - 4x2y + 2xy2 + 4x2y - 2xy2 + y3
= 8x3 + y3.
Cho \(\dfrac{\text{x}}{\text{3}}=\dfrac{\text{y}}{\text{4}}\) . Tính x,y biết :
A) x+y=35
B) 2x+y=50
C) y-x = 3
D) x . y = 108
tính giá trị của biểu thức
a) \(A=2x^2-\dfrac{1}{3}y,t\text{ại}x=2;y=9\)
b) \(P=2x^2+3xy+y^2t\text{ại }x=-\dfrac{1}{2};y=\dfrac{2}{3}\)
c) \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)t\text{ại}x=2;y=\dfrac{1}{4}\)
Cho x + 3y - 2z = 36. Tìm x,y,z biết
a) \(\dfrac{\text{x-1}}{\text{3}}=\dfrac{\text{y+2}}{\text{4}}=\dfrac{\text{z-2}}{\text{3}}\)
b) \(\dfrac{\text{x}}{\text{4}}=\dfrac{\text{y}}{\text{3}};\dfrac{\text{y}}{\text{2}}=\dfrac{\text{z}}{\text{5}}\)
c) 9x = 5y ; 2x = z
d) 2x = 3y = 4z
Cho x + 3y - 2z = 36 . Tìm x,y,z biết :
a)\(\dfrac{\text{x-1}}{\text{3}}=\dfrac{\text{y+2}}{\text{4}}=\dfrac{\text{z-2}}{\text{3}}\)
b)\(\dfrac{\text{x}}{\text{4}}=\dfrac{\text{y}}{3};\dfrac{\text{y}}{\text{2}}=\dfrac{\text{z}}{\text{5}}\)
c) 9x = 5y ; 2x = z
d) 2x = 3y = 4z
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
1/ x\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\text{và}2x+3y-z=50\)
2/ x : y : z = 3 : 5 ; ( - 2 ) và 5x - y + 3z = -16
3/ 2x + 3y ; 7z = 5y và 3x - 7y + 5z = 30
4/ \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\text{và}x-y-z=38\)
a)\(\frac{z}{5}=\frac{x}{2}=\frac{y}{3}v\text{à}x.y-z=810\)
b)\(5x=3yv\text{à}2x^2-y^2=-28\)
c)\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}v\text{à}x^2+y^2+z^2=14\)
d)\(x:y:z=3:4:5v\text{à}5z^2-2y^2=594\)
Cho \(\frac{x^{\text{4}}}{a}+\frac{y^{\text{4}}}{b}=\frac{1}{a+b};x^2+y^2=1\)
Chứng minh rằng:\(\frac{x^{200\text{4}}}{a^{1002}}+\frac{y^{200\text{4}}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
Cho \(\dfrac{\text{x}}{\text{2}}=\dfrac{\text{y}}{\text{3}}=\dfrac{\text{z}}{\text{5}}\). Tìm x,y,z biết
a) x + y + z = 40
b) x - 3y + 2z = 9
c) x -y + z = 28
d) 3x + 2y = 24