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a) \(\dfrac{6x^2y^3-2x^2y+6xy}{6xy}\)
\(=\dfrac{6x^2y^3}{6xy}-\dfrac{2x^2y}{6xy}+\dfrac{6xy}{6xy}\)
\(=xy^2-\dfrac{x}{3}+1\)
b) \(\dfrac{4\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{2\left(x+y\right).2\left(x+y\right)^2}{2\left(x+y\right)}\)
\(=2\left(x+y\right)^2\)
c) \(\dfrac{8x^3+27y^3}{2x+3y}\)
\(=\dfrac{\left(2x\right)^3+\left(3y\right)^3}{2x+3y}\)
\(=\dfrac{\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]}{2x+3y}\)
\(=4x^2-6xy+9y^2\)
d) \(\dfrac{48x^4y^3-12x^2y^5+6x^2y^2}{3x^2y^2}\)
\(=\dfrac{48x^4y^3}{3x^2y^2}-\dfrac{12x^2y^5}{3x^2y^2}+\dfrac{6x^2y^2}{3x^2y^2}\)
\(=16x^2y-4y^3+2\)
a) Ta có: \(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)
\(=\left(x+4\right)^3\)
b) Ta có: \(x^3-12x^2+48x-64\)
\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)
\(=\left(x-4\right)^3\)
c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=\left(2x+y\right)^3\)
d)Sửa đề: \(x^3-3x^2+3x-1\)
Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
e) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
a: Sửa đề: \(-x^3-12x^2-48x-64\)
\(=-\left(x+4\right)^3\)
\(=-\left(-6+4\right)^3=-\left(-2\right)^3=-\left(-8\right)=8\)
b: \(=8x^3-y^3-8x^3+27y^3=26y^3=26\cdot\left(-3\right)^3=-702\)
c: \(=-\left(4x^4-12x^2y+9y^2\right)\)
\(=-\left(2x^2-3y\right)^2\)
\(=-\left(2x^2-2x-11\right)^2\)
A = x3 + 3x2 + 3x - 899
= (x3 + 3x2 + 3x + 1) - 900
= (x + 1)3 - 900
= (29 + 1)3 - 900 = 303 - 900 = 26100
B = x3 - 6x2 + 12x + 10
= (x3 - 6x2 + 12x - 8) + 18
= (x - 2)3 + 18
= (12 - 2)3 + 18 = 103 + 18 = 1000 + 18 = 1018
c) C = 8x3 - 27y3
= (2x)3 - (3y)3
= (2x - 3y)(4x2 + 6xy + 9y2)
= (2x - 3y)(4x2 - 12xy + 9y2) + (2x - 3y).18xy
= (2x - 3y)(2x - 3y)2 + (2x - 3y).18xy
= (2x - 3y)3 + (2x - 3y).18xy
= 53 + 5.18.4
= 125 - 360
= -235
D = x3 + y3 + 3xy(x2 + y2) + 6x2y2(x + y)
= (x + y)(x2 - xy + y2) + 3x3y + 3xy3 + 6x2y2
= x2 + y2 - xy + 3x3y + 3xy3 + 6x2y2
= (x + y)2 - 3xy + 3x3y + 3xy3 + 6x2y2
= 1 - 3xy(2xy - 1) + 3xy(x2 + y2)
= 1 - 3xy(x2 + y2 + 2xy - 1)
= 1 - 3xy[(x + y)2 - 1]
= 1 - 0 = 1
Bài 1 : Khai triển :
a, \(\left(x+5\right)^2=x^2+10x+25\)
b, \(\left(x-3y\right)^2=x^2-6xy+9y^2\)
c, \(\left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\)
d, \(\left(x+3y\right)^3=x^3+9x^2y+27xy^2+27y^3\)
e, \(27x^3-9y^2+y-\frac{1}{27}=\left(3x-\frac{1}{3}\right)^3\)
g, \(8x^6+12x^4y+6x^2y^2+y^3=\left(2x^2+y\right)\)
h, \(4x^2+12x^4y+6x^22y^2+y^3=\left(\sqrt[3]{4x^2}+y\right)\)
1)
2x^2 + 12 + 2x.(4-x) = 0
2x^2 + 12 + 8x - 2x^2 = 0
12 + 8x = 0
8x = -12
x = -4/3
2)
A= 8x3 - 12x2y +6xy2 - y3
A = (2x)3 - 3.(2x)2.y + 3.2.x.y2 - y3
A = (2x-y)3
...
tính giá trị mk nhường bn làm đó!
Bn lam dk chua Anh