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a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
a) x^2+2xy+y^2-16
=(x+y)2-16
=(x+y-4)(x+y+4)
b) 3x^2+5x-3xy-5y
=(3x2-3xy)+(5x-5y)
=3x(x-y)+5(x-y)
=(x-y)(3x+5)
c) 4x^2-6x^3y-2x^2+8x
ko bik hoặc sai đề
d) x^2-4-2xy+y^2
=(x-y)2-4
=(x-y+2)(x-y-2)
e) x^3-4x^2-12x+27
=sai đề
g) 3x^2-18x+27
=3(x2-6x+9)
=3(x-3)2
h) x^2-y^2-z^2-2yz
=x2-(y2+z2+2yx)
=x2-(y+z)2
=(x-y-z)(x+y+z)
k) 4x^2(x-6)+9y^2(6-x)
=4x2(x-6)-9y2(x-6)
=(x-6)(4x2-9y2)
=(x-6)(2x-3y)(2x+3y)
l)6xy+5x-5y-3x^2-3y^2
=(5x-5y)+(-3x2+6xy-3y2)
=5(x-y)-3(x2-2xy+y2)
=5(x-y)-3(x-y)2
=(x-y)(5-3(x-y))
=(x-y)(5-3x+3y)
a) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y\right)^3-3\left(2x^3y\right)^20,5x^2+3.2x^3y\left(0,5x^2\right)^2-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+1,5x^7y-0,125x^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=x^3-3^3\)
\(=x^3-27.\)
\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)
a,\(\left(2x^3y-0,5x^2\right)^3=\left(2x^3y\right)^3-3.\left(2x^3y\right)^2.\left(0,5x^2\right)+3.\left(0,5x^2\right)^2.\left(2x^3y\right)-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\frac{3}{2}x^7y-\frac{1}{8}x^6\)
b,\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)
\(=\left(x^2\right)^3-3^3=x^6-27\)
\(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^2+3x^2y+9xy^2-3x^2y-9xy^2-27y^3\)
\(=\)\(x^2-27y^3\)
\(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3xy}+4y^2\right)\)
\(=\)\(\frac{x^3}{27}-\frac{2}{9xy}+\frac{4xy^2}{3}+\frac{2x^2y}{9}-\frac{4y}{3xy}+8y^3\)
làm nốt nha
\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\)
\(=x^4+\frac{2}{5}x^2y-\frac{2}{5}x^2y-\left(\frac{2}{5}y\right)^2\)
\(=x^4-\frac{4}{25}y^2\)
a/ \(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)
b/ (x - 3y)(x2 + 3xy + 9y2) = x3 - 27y3
c/ (5 + 3x)3 = 125 + 225x + 135x2 + 27x3
a) ( x^2 + 2/5y).(x^2-2/5y)
= x^4 - 4/25y^2
b) ( x- 3y)( x^2 + 3xy + 9y^2)
= x^3 - 27y^3
c) (5+3x)^3
= 125 + 225x + 135x^3 + 27x^3
a)\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)=x^4-\frac{4}{25}y\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
c) \(\left(5+3x\right)^3=5^3+3.5^2.3x+3.5.\left(3x\right)^2+\left(3x\right)^3=125+225x+135x^2+27x^3\)
Vi Dương: tớ cx giống cậu .
\(a,x^4-\frac{4}{25}y^2\)
\(b,x^3-27y^3\)
\(c,125+45x+27x^3\)
Ta có các hằng đẳng thức tương ứng :
a/ (a - b)(a + b) = a2 - b2
b/ (a - b)(a2 + ab + b2) = a3 - b3
c/ (a + b)3 = a3 + 3a2b + 3ab2 + b3
Bạn cứ dựa vào các hằng đẳng thức này mà làm ra nha
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-3x^2y+3x^2y-9xy^2+9xy^2-27y^3\)
\(=x^3-27y^3\)
a/ (a - b)(a + b) = a2 - b2
Áp dụng hđt trên ta được;
\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)
b/ (a - b)(a2 + ab + b2) = a3 - b3
Áp dụng hđt trên:
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+3xy+\left(3y\right)^2\right]=x^3-\left(3y\right)^3=x^3-27y^3\)
c/ (a + b)3 = a3 + 3a2b + 3ab2 + b3
Áp dụng bđt trên:
\(\left(5+3x\right)^3=5^3+3.5^2.3x+3.5.\left(3x\right)^2+\left(3x\right)^3\)
\(=125+225x+135x^2+27x^3\)
\(\left(5+3x\right)^3\)
\(=\left(5+3x\right)\left(5+3x\right)\left(5+3x\right)\)
\(=\left(25+15x+15x+9x^2\right)\left(5+3x\right)\)
\(=125+75x+75x+45x^2+75x+45x^2+45x^2+27x^3\)
\(=125+\left(75x+75x+75x\right)+\left(45x^2+45x^2+45x^2\right)+27x^3\)
\(=125+\left(75+75+75\right)x+\left(45+45+45\right)x^2+27x^3\)
\(=125+225x+135x^2+27x^3\)
A)=x4-4/25y2
B)=x3-27y3
c)=27x3+135x2+225x+125
Chuc ban học tot nha nho duyệt cho min nua nha!