Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) (x^2+2xy+y^2) : (x+y)
=(x+y)2:(x+y)
=x+y
b) (125x^3+1) : (5x+1)
=(5x+1)(25x2-5x+1):(5x+1)
=25x2-5x+1
c) (x^2-2xy+y^2) : (y-x)
=(x-y)2:(y-x)
=-(x-y)2:(x-y)
=-(x-y)
=-x+y
a) (x2 + 2xy + y2) : (x + y);
=(x+y)2:(x+y)
=x+y
b) (125x3 + 1) : (5x + 1);
=(5x+1)(25x2-5x+1):(5x+1)
=25x2-5x+1
c) (x2 – 2xy + y2) : (y – x).
=(x-y)2:(y-x)
=(y-x)2:(y-x)
=y-x
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
Bài giải:
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\)
\(=\left(x+y\right)^2:\left(x+y\right)\)
\(=x+y\)
b) \(\left(125x^3+1\right):\left(5x+1\right)\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)\)
\(=25x^2-5x+1\)
c) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
a ) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\)
\(=\left(x+y\right)^2:\left(x+y\right)\)
\(=\left(x+y\right)\)
b ) \(\left(125x^3+1\right)\left(5x+1\right)\)
\(=\left[\left(5x\right)^3+1\right]:\left(5x+1\right)\)
\(=\left(5x\right)^2-5x+1\)
\(=25x^2-5x+1\)
c ) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left[-\left(x-y\right)\right]\)
\(=-\left(x-y\right)\)
\(=y-x\)
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\\ =\left(x+y\right)^2:\left(x+y\right)\\ =\left(x+y\right)\)
b) \(\left(125x^3+1\right)\left(5x+1\right)\\=\left[\left(5x\right)^3+1\right]:\left(5x+1\right)\\ =\left(5x\right)^2-5x+1 \\ =25x^2-5x+1\)
c) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\\ =\left(x-y\right)^2:\left[-\left(x-y\right)\right]\\ =-\left(x-y\right)\\ =y-x\)
Bài 3 :
a )
\(4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x=0\) or \(x=1\)
b )
|
|||||||||||||||||||||||||||||
1.
a.\(4x^2\left(5x^3-3x+1\right)\)
\(=20x^5-12x^3+4x^2\)
b.\(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
c.\(\left(x^2-2xy+y^2\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\left(x-y\right)\)
\(=\left(x-y\right)^3\)
2.
a.Bạn xem lại đề câu này nhé!
b.\(x^2-y^2-3x-3y\)
\(=\left(x^2-y^2\right)+\left(-3x-3y\right)\)
\(=\left(x+y\right)\left(x-y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-3\right)\)
3.
a.\(4x^2-4x=0\)
\(4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy x=0 hoặc x=1.
\(\frac{x^2+2xy+y^2}{x+y}=\frac{\left(x+y\right)^2}{x+y}=x+y\)
\(\frac{125x^3+1}{5x+1}=\frac{\left(5x\right)^3+1}{5x+1}=\frac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)
\(\frac{2x^3+5x^2-2x+3}{2x^2-x+1}=\frac{\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)}{2x^2-x+1}\)
\(=\frac{x\left(2x^2-x+1\right)+3.\left(2x^2-x+1\right)}{2x^2-x+1}=\frac{\left(2x^2-x+1\right)\left(x+3\right)}{2x^2-x+1}=x+3\)
Tham khảo nhé~
a, \(\left(x^2+2xy+y^2\right):\left(x+y\right)=\left(x+y\right)^2:\left(x+y\right)=x+y\)
b, \(\left(125x^3+1\right):\left(5x+1\right)=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)=25x^2-5x+1\)
c, \(\left(x^2-2xy+y^2\right):\left(y-x\right)=\left(x-y\right)^2:\left(y-x\right)=\left(y-x\right)^2:\left(y-x\right)=y-x\)
kick mình nha