Tính hằng đẳng thức đáng nhớ:
a) (x + 4 )(x2- 4x + 16)
b) (x-3y)(x2+3xy+9y2)
c) (1/3x + 2y ) ( 1/9x2- 2/3xy + 4y2)
d) (x2-1/3)(x4 + 1/3x2 + 1/9)
TÍNH GIÚP EM VS Ạ. THANKS
(x+4)(x2-4x+16)
(x-3y)(x2+3xy+9y2)
(x2-\(\dfrac{1}{3}\))(x4+\(\dfrac{1}{3}\)x2+\(\dfrac{1}{9}\))
\(=x^3+64\\ =x^3-27y^3\\ =x^6-\dfrac{1}{27}\)
\(\left(x+4\right)\left(x^2-4x+16\right)=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
viết các biểu thức sau dưới dạng tổng hoặc hiệu của 2 lập phương:
a, (3x - 1) (9x2 + 3x + 1)
b, (1 - \(\dfrac{x}{5}\)) (\(\dfrac{x^2}{25}\) + \(\dfrac{x}{5}\) + 1)
c, (x +3y) (x2 - 3xy + 9y2)
d, (4x + 3y) (16x2 - 12xy + 9y2)
a: \(\left(3x-1\right)\left(9x^2+3x+1\right)=27x^3-1\)
b: \(\left(1-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{5}+1\right)=1-\dfrac{x^3}{125}\)
c: \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
d: \(\left(4x+3y\right)\left(16x^2-12xy+9y^2\right)=64x^3+27y^3\)
Bài 5 .Tính b ằng cách hợp lí:
a) 193 b) 2013
Bài 6.Đơn giản biểu thức:
a) (x-2)(x2+2x+4)
c) (x-3y)(x2+3xy+9y2)
d) (x2-3)(x4+3x2+9)
\(5,\\ a,19^3=\left(20-1\right)^3=8000-3\cdot400\cdot1+3\cdot20\cdot1-1\\ =8000-1200+60-1=6859\\ b,201^3=\left(200+1\right)^3=8000000+3\cdot40000\cdot1+3\cdot200\cdot1+1\\ =8000000+120000+600+1=8120601\)
chứng tỏ
a) x2 + 8y2 =( x +2y ) ( x2- 2xy +4y2)
b) (x-y) (x2+xy+y2 ) -3xy (x-y) =( x-y)3
c) (x-3y) (x2 +3xy +9y2 ) - ( 3y +x ) ( 9y2 -3xy + x2) = -54y3
cíu em vớii
\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)
\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)
Tính giá trị của biểu thức:
a) N= (25x2 + 10xy + 4y2)(5x - 2y) tại x=1/5:y=1/2
b) Q= (x + 3y)(x2 - 3xy + 9y2) tại x=y=1/2
Giúp mik với ạ!
a: \(N=\left(5x\right)^3-\left(2y\right)^3=1^3-1^3=0\)
b: \(Q=x^3+27y^3=\dfrac{1}{8}+\dfrac{27}{8}=\dfrac{28}{8}=\dfrac{7}{2}\)
bài 2: viết cá đa thức sau dưới dạng hằng đẳng thức đáng nhớ sau :
a,x2+2x+1=
b,y2+4y+4=
c,9-6x+x2=
d,a2-14a+49=
e,m2-4m+4=
f,4x2-4x+1=
g,a2+10a+25=
h,100-20z+z2=
i,x2+6xy+9y2=
j,4x2-12xz+25b2=
k,a2+10ab+25b2=
l,x4+2x2+1=
m,y6-2y3+1=
n,c10-10c5+25=
o,9x4+12x2y+4y2=
p,25m4n6-10m2n3=
em đang cần gấp ,giúp em với
\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)
Tính:
a) (\(\dfrac{1}{3}\)x+2y).(\(\dfrac{1}{9}\)x2-\(\dfrac{2}{3}\)xy+4y2)
b) (x2-\(\dfrac{1}{3}\)).(x4+\(\dfrac{1}{3}\)x2+\(\dfrac{1}{9}\))
c) (y-5).(25+5y+y2+2y)
d) (5x+3y).(25x2-15xy+9y2)
Giải chi tiết giúp mình nha.Cảm ơn
a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
a) (2x + 3y)2
b) (x + \(\dfrac{1}{4}\))2
c) (x2 + \(\dfrac{2}{5}\)y) . (x2 - \(\dfrac{2}{5}\)y)
d) (2x + y2)3
e) (3x2 - 2y)2
f) (x + 4) (x2 - 4x + 16)
g) (x2 - \(\dfrac{1}{3}\)) . (x4 + \(\dfrac{1}{3}\)x2 + \(\dfrac{1}{9}\))
a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)
d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)
f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)