Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 16x2(x - y)2 - 10y(y - x)3
= 16x2(y - x)2 - 10y(y - x)3
= 2(y - x)2[8x2 - 5y(y - x)]
= 2(y - x)2(8x2 + 5xy - 5y2)
b) a2 -b2 + 4ab - 9 (sai đề)
a) \(2a^{n+2}b^n-18a^nb^{n+2}\)
\(=2a^nb^n\left(a^2-9b^2\right)\)
\(=2a^nb^n\left(a-3b\right)\left(a+3b\right)\)
\(\left(2a+b\right)^2-\left(2a+a\right)^2\)
\(=\left(2a+b-2a-a\right)\left(2a+b+2a+a\right)\)
\(=\left(b-a\right)\left(5a+b\right)\)
\(\left(2a+b\right)^2-\left(2a+a\right)^2\)
\(=\left(2a+b\right)^2-\left(3a\right)^2\)
\(=\left(2a+b-3a\right)\left(2a+b+3a\right)\)
\(=\left(b-a\right)\left(5a+b\right)\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)