Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(x-2y=-4\Rightarrow\left(x-2y\right)^2=16\)
\(\Rightarrow x^2-4xy+4y^2=16\Rightarrow x^2+4y^2=16+4xy=16+4.6=40\)
\(x^3-8y^3=\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(-4\right)\left(40+2.6\right)=-208\)
b) Ta có: \(x+3y=10\Rightarrow x^2+6xy+9y^2=100\Rightarrow x^2+9y^2=100-6xy=100-6.3=82\)
\(x^3+27y^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=10\left(82-3.3\right)=730\)
\(x^3+27y^3=1-9xy\left(x+3y\right)\)
<=> \(x^3+27y^3+9xy\left(x+3y\right)=1\)
<=> \(\left(x+3y\right)^3=1\)
<=> \(x+3y=1\)
Vậy \(M=1\)
\(x^3+27x^3=1-9xy\left(x+3y\right)\))
\(=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=1-9xy\left(x+3y\right)\)
=\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)-1+9xy\left(x+3y\right)=0\)
=\(\left(x+3y\right)\left(x^2-3xy+9y^2+9xy\right)-1=0\)
=\(\left(x+3y\right)\left(x^2+6xy+9y^2\right)-1=0\)
=\(\left(x+3y\right)\left(x+3y\right)^2-1=0\)
=\(\left(x+3y\right)\left(x+3y\right)^2=1\)
\(\Rightarrow x+3y=\left(x+3y\right)^2=1\)
\(\Rightarrow x+3y=1\)
\(1.\)
\(a.\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-3^3\right)-\left(54+x^3\right)\)
\(=x^3-27-54-x^3\)
\(=-81\)
\(b.\)
\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(27x^3+y^3\right)-\left(27x^3-y^3\right)\)
\(=27x^3+y^3-27x^3+y^3\)
\(=2y^3\)
\(2.\)
\(a.\)
\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
\(b.\)
\(\left(2x-3y\right)\left(4x^2+6xy+9y^3\right)=8x^3-27y^3\)
1) a) \(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-3^3\right)-\left(54+x^3\right)\\ =\left(x^3-27\right)-54-x^3\\ =-27-54\\ =-81\)
b) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left[\left(3x\right)^3+y^3\right]-\left[\left(3x\right)^3-y^3\right]\\ =2y^3\)
2) a) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
b) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)
a: \(=-8x^5+6x^3-2\)
b: \(=-\dfrac{2}{3}x+7-x^2y\)
c: \(=\dfrac{7\left(x-y\right)^4+4\left(x-y\right)^3}{\left(x-y\right)^2}=7\left(x-y\right)^2+4\left(x-y\right)\)
d: \(=\dfrac{6\left(x-3y\right)^4}{5\left(x-3y\right)}=\dfrac{6}{5}\left(x-3y\right)^3\)
a)(x+3y)(x2-3xy+9y2)=x3+27y3
b)(2x-3y)(4x2+6xy+9y2)=8x3-27y3
Bạn gõ latex đc ko, cái này này chứ đề khó hiểu quá
Nhấn vào chữ"tex" ở kia kìa nó là mã latex đấy