Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Sửa đề: y=2/3
\(P=\left(3y+\dfrac{1}{3}\right)^3=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)
b: \(Q=x^2+4xy+4y^2-2x-4y+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2\cdot5+10=25\)
a: Sửa đề: \(P=27y^3+9y^2+y+\dfrac{1}{27}\)
\(=\left(3y+\dfrac{1}{3}\right)^3\)
\(=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)
b: \(Q=x^2+4xy+4y^2-2x-4y+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=25-2\cdot5+10=25\)
a: \(P=\left(3y+\dfrac{1}{3}\right)^3=\left(3\cdot\dfrac{2}{3}+\dfrac{1}{3}\right)^3=\left(\dfrac{7}{3}\right)^3=\dfrac{343}{27}\)
b: \(Q=x^2+4xy+4y^2-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
=25
\(P=27y^3+9y^2+y+\dfrac{1}{27}=\left(3y+3\right)^3\)
Với \(y=\dfrac{2}{3}\) ta có:
\(P=\left(3.\dfrac{2}{3}+3\right)^3=5^3=125\)
\(Q=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2-2x+4xy\right)+4y^2-4y+10\)
\(=\left[x^2-2x\left(1-2y\right)+\left(1-2y\right)^2\right]+4y^2-4y+10-\left(1-2y\right)^2\)\(=\left(x+2y-1\right)^2+4y^2-4y+10-1+4y-4y^2\)\(=\left(x+2y-1\right)^2+9\)
Với \(x+2y=5\) , ta có:
\(Q=\left(5-1\right)^2+9=25\)
a: \(\dfrac{1}{6x^2y^3}=\dfrac{7x^2}{42x^4y^3}\)
\(\dfrac{-5}{21xy^2}=\dfrac{-10x^3y}{42x^4y^3}\)
\(\dfrac{3}{14x^4y}=\dfrac{3\cdot3y}{42x^4y^3}=\dfrac{9y}{42x^4y^3}\)
b: \(\dfrac{2}{x^3-y^3}=\dfrac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{2x+1}{x^2-y^2}=\dfrac{\left(2x+1\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)
\(=\left(2x\right)^3-1=8x^3-1\)
b,\(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)
`a)(2x-1)(4x^2+2x+1)`
`=(2x-1)[(2x)^2+2x.1+1^2]`
`=(2x)^3-1^3`
`=8x^3-1`
Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`
`b)(x+2y+z)(x+2y-z)`
`=[(x+2y)+z][(x+2y)-z]`
`=(x+2y)^2-z^2`
`=x^2+2.x.2y+(2y)^2-z^2`
`=x^2+4xy+4y^2-z^2`
Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`
`(A+B)^2=A^2+2AB+B^2`
-(a - 3)2 = -(a2 - 6a + 9) = -a2 + 6a - 9
(x - 2)(x + 2) = x2 - 4
-(5 + 4y)(5 - 4y) = -(25 - 16y2) = -25 + 16y2
(\(\dfrac{1}{2}\)x + 2y)(\(\dfrac{1}{2}\)x - 2y) = \(\dfrac{1}{4}\)x2 - 4y2