Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(x^2-2xy+y^2-16\)
\(=\left(x-y\right)^2-16\)
\(=\left(x-y-4\right)\left(x-y+4\right)\)
p/s: chúc bạn học tốt
\(x^2-2xy+y^2-16\)
\(\Rightarrow\left(x-y\right)^2-16\)
\(\Rightarrow\left(x-y-4\right)\left(x-y+4\right)\)
Code : Breacker
\(x^2\left(x-1\right)+16\left(1-x\right)\)
\(=x^2\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4^2\right)\)
\(=\left(x-1\right)\left(x-4\right)\left(x-4\right)\)
a)
\(2x^2y-8xy^2\\ =2xy\left(x-4y\right)\)
b)
\(x^2-2xy+y^2-16\\ =\left(x^2-2xy+y^2\right)-16\\ =\left(x-y\right)^2-16\\ =\left(x-y-4\right)\left(x-y+4\right)\)
a) 5x2 -20
= 5(x2 -4)
=5 (x2 -22)
= 5(x-2)(x+2)
b) 16 - (x+y)2
=42 -(x+y)2
= (4-x-y)(4+x+y)
a, \(5\left(x^2-4\right)=5\left(x-2\right)\left(x+2\right)\)
b, \(16-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)
mấy bài này áp dụng hđt là được nhé
\(\left(x+3\right)^2-16\)
\(=\left(x+3-4\right)\left(x+3+4\right)\)
\(=\left(x-1\right)\left(x+7\right)\)
x 16 + x 8 − 2 = ( x 8 ) 2 + x 8 − 2 = ( x 8 − 1 ) ( x 8 + 2 ) = ( x 4 − 1 ) ( x 4 + 1 ) ( x 8 + 2 ) = ( x 2 − 1 ) ( x 2 + 1 ) ( x 4 + 1 ) ( x 8 + 2 ) = ( x − 1 ) ( x + 1 ) ( x 2 + 1 ) ( x 4 + 1 ) ( x 8 + 2 )
\(\left(x+2\right)^2-16\\ \backslash=\left(x+2-4\right)\left(x+2+4\right)\\ =\left(x-2\right)\left(x+6\right)\)
(x+2)2-42=(x+2-4)(x+2+4)