(-x2 y)3. \(\dfrac{1}{2}\)x2y3.(-2xy2z)2
thu gọn đa thức và tìm bậc
3x4y2 + \(\dfrac{\text{1}}{\text{2}}\) xy5 - \(\dfrac{\text{3}}{\text{4}}\) x2y3 -
\(A=\left(3x^4y^2-\dfrac{1}{2}x^4y^2\right)+\left(\dfrac{1}{2}xy^5+5xy^5\right)+\left(\dfrac{-3}{4}x^2y^3-\dfrac{1}{4}x^2y^3\right)=\dfrac{11}{4}x^4y^2+\dfrac{26}{5}xy^5-x^2y^3\)
Bậc là 6
Thực hiện phép tính và tìm hệ số; biến; bậc của đơn thức thu được
a) (2x2y3).(-\(\dfrac{5}{2}\)x2y3)
b) (6x2y2z).(\(\dfrac{1}{3}\)xy3)
c) 8xy2+5xy2-4xy2
d) -\(\dfrac{1}{2}\)x2y+\(\dfrac{1}{3}\)x2y-x2y
a) -5x4y6
Hệ số là: -5
biến là x4y6
Bậc là 10
2 x3y5z
Hệ số là 2
Biến là x3y5x
bậc là 9
-160x3y6
Hệ số là : -160
Biến là x3y6
Bậc là:9
-1/6x6y3
hệ số là -1/6
Biến là x6y3
bậc là 9
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) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
Bài 3: Trong các biểu thức sau, đâu là đơn thức?
(1-\(\dfrac{1}{\sqrt{3}}\)) x2; \(\dfrac{1}{2}\)(x2 - 1); x2. \(\dfrac{7}{2}\); 6\(\sqrt{y}\); \(\dfrac{1-\sqrt{5}}{x}\); \(\dfrac{x-y^2}{4}\)
Các đơn thức là :
\(\left(1-\dfrac{1}{\sqrt[]{3}}\right)x^2;x^2.\dfrac{7}{2}\)
A = x2y3 + xy tại x = - 1 và y = 2.
B = 2x2 + y4 + xyz – 5 tại x = 3 ; y = 2; z = 1.
a: \(A=\left(-1\right)^2\cdot2^3+\left(-1\right)\cdot2=8-2=6\)
b: \(B=2\cdot2^2+2^4+3\cdot2\cdot1-5=8+16+6-5=8+16+1=25\)
Thu gọn các đa thức sau
a: A = -2xy + 3\2xy2 + 1\2 xy2 + xy
b: B = xy2z + 2xy2z -xyz -3xy2z + xy2z
c: C = 4x2y3 + x4 -2x2 + 6x4 -x2y3
d: D = 3\4xy2 - 2xy - 1\2xy2 + 3xy
e: E = 2x2 - 3y3 - z4 - 4x2 + 2y3 + 3z4
f: F = 3xy2z +xy2z - xyz + 2xy2z - 3xyz
a: A = -2xy + 3/2xy^2 + 1/2xy^2 + xy = -2xy + 2xy^2 + xy = 2xy^2 - xy
b: B = xy^2z + 2xy^2z - xyz - 3xy^2z + xy^2z = 3xy^2z - xyz
c: C = 4x^2y^3 + x^4 - 2x^2 + 6x^4 - x^2y^3 = 7x^4 + 3x^2y^3 - 2x^2
d: D = 3/4xy^2 - 2xy - 1/2xy^2 + 3xy = 5/4xy^2 + xy
e: E = 2x^2 - 3y^3 - z^4 - 4x^2 + 2y^3 + 3z^4 = -2x^2 - y^3 + 2z^4
f: F = 3xy^2z + xy^2z - xyz + 2xy^2z - 3xyz = 6xy^2z - 2xyz
a: A=-2xy+3/2xy^2+1/2xy^2+xy
=-2xy+xy+3/2xy^2+1/2xy^2
=2xy^2-xy
b: \(B=xy^2z+2xy^2z-xyz-3xy^2z+xy^2z\)
\(=xy^2z\left(1+2-3+1\right)-xyz=xy^2z-xyz\)
c: \(=4x^2y^3-x^2y^3+x^4+6x^4-2x^2\)
\(=7x^4-x^2+3x^2y^3\)
d: \(=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+3xy-2xy\)
=1/4xy^2+xy
e: \(=2x^2-4x^2-3y^3+2y^3+3z^4-z^4\)
\(=-2x^2-y^3+2z^4\)
f: \(=xy^2z+3xy^2z+2xy^2z-xyz-3xyz\)
\(=6xy^2z-4xyz\)
(\(\dfrac{2}{3}\)x2-\(\dfrac{1}{2}\)y)3
\(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\dfrac{8}{27}x^6-\dfrac{2}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)
(3x2-2y)3
(\(\dfrac{2}{3}\)x2-\(\dfrac{1}{2}\)y)3
(2x+\(\dfrac{1}{2}\))3
(x-3)3
\(\left(x-3\right)^3=x^3-9x^2+27x-27\)
\(\left(2x+\dfrac{1}{2}\right)^3=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)
1. Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng :
a) x2y3/5 = 7x3y4/35xy
b) x3 - 4x/10-5x = -x2-2x/5
c)x + 2/ x-1 = (x+2)(x+1)/ x2-1
d) x2 - x - 2/ x+1 = x2 - 3x +2/ x-1
e) x3+8/ x2-2x+4 = x+2
a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)
b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{-x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)
c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+2}{x-1}\)
d: \(\left(x^2-x-2\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2-3x+2\right)\left(x+1\right)\)
=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)